Punjab Board Class 11 Mathematics

Chapter 14: Vectors in Space — Class 11 Maths Notes

Vectors in space: dot and cross products, lines and planes in 3D with geometric meaning made clear. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 14.

Exercise solutions — Chapter 14

Every exercise of Vectors in Space solved step by step.

Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4

Mathematics 11 (Punjab Board)

Chapter 14: Vectors in Space — Theory

14.1 Vectors in Space

A vector in space is written with three components using the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) along the \(x\)-, \(y\)- and \(z\)-axes:

\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} = [u_1, u_2, u_3]

Addition, subtraction aur scalar multiplication component-wise hoti hai. Do vectors parallel hote hain agar ek doosre ka scalar multiple ho: \(\mathbf{v} = c\mathbf{u}\).

Key point: Every vector operation in space works component by component — exactly like 2D, with a third component added.

Magnitude, Distance and Direction Cosines

The magnitude (length) of \(\mathbf{u} = [u_1, u_2, u_3]\) and the distance between \(P_1(x_1,y_1,z_1)\) and \(P_2(x_2,y_2,z_2)\) are:

|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}, \qquad d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

Agar vector \(x, y, z\) axes ke sath angles \(\alpha, \beta, \gamma\) banaye to direction cosines hain:

\cos \alpha = \frac{u_1}{|\mathbf{u}|}, \quad \cos \beta = \frac{u_2}{|\mathbf{u}|}, \quad \cos \gamma = \frac{u_3}{|\mathbf{u}|}, \qquad \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

A unit vector in the direction of \(\mathbf{u}\) is \(\hat{\mathbf{u}} = \frac{\mathbf{u}}{|\mathbf{u}|}\).

Key point: Direction cosines always satisfy \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\) — this one identity decides many MCQs.

14.2 Dot (Scalar) Product

The dot product of \(\mathbf{u} = [u_1,u_2,u_3]\) and \(\mathbf{v} = [v_1,v_2,v_3]\) can be computed from components or from the angle \(\theta\) between them:

\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 = |\mathbf{u}|\,|\mathbf{v}|\cos\theta

Is se angle ka formula milta hai: \(\cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\). Aur \(\mathbf{a}\) ki \(\mathbf{b}\) par projection (scalar) hai \(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\).

Key point: \(\mathbf{u} \cdot \mathbf{v} = 0\) (for non-zero vectors) is the test for perpendicularity. Work done is also a dot product: \(W = \mathbf{F} \cdot \mathbf{d}\).

14.3 Cross (Vector) Product

The cross product \(\mathbf{u} \times \mathbf{v}\) is a vector perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\), with magnitude \(|\mathbf{u}||\mathbf{v}|\sin\theta\). Component form mein ye determinant hai:

\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

Key properties: \(\mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u})\) (anti-commutative), \(\mathbf{u} \times \mathbf{u} = \mathbf{0}\), aur \(\mathbf{u} \times \mathbf{v} = \mathbf{0}\) exactly when \(\mathbf{u}\) and \(\mathbf{v}\) are parallel.

Key point: \(\mathbf{u} \times \mathbf{v} = \mathbf{0}\) is the test for parallel vectors; its magnitude gives areas — \(|\mathbf{u} \times \mathbf{v}|\) for a parallelogram, \(\frac{1}{2}|\mathbf{u} \times \mathbf{v}|\) for a triangle.

Angle via Cross Product and Moment of a Force

Sine of the angle between two vectors:

\sin\theta = \frac{|\mathbf{u} \times \mathbf{v}|}{|\mathbf{u}|\,|\mathbf{v}|}

A unit vector perpendicular to the plane of \(\mathbf{u}\) and \(\mathbf{v}\) is \(\frac{\mathbf{u} \times \mathbf{v}}{|\mathbf{u} \times \mathbf{v}|}\). The moment (torque) of a force \(\mathbf{F}\) applied at position \(\mathbf{r}\) is \(\mathbf{M} = \mathbf{r} \times \mathbf{F}\).

Key point: Cross product gives a perpendicular vector; dot product gives a number. Moment = position cross force.

14.4 Scalar Triple Product

The scalar triple product \([\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\) is a number, computed as a \(3 \times 3\) determinant:

\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}

Cyclic order mein vectors interchange karne se value nahi badalti: \([\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = [\mathbf{v}\;\mathbf{w}\;\mathbf{u}] = [\mathbf{w}\;\mathbf{u}\;\mathbf{v}]\). Agar do rows same hon to value zero.

Key point: Triple product is a determinant — expand along the first row and compute carefully; one sign error changes everything.

Volumes and Coplanarity

The absolute value of the scalar triple product is the volume of the parallelepiped formed by the three vectors; one-sixth of it is the volume of the tetrahedron:

V_{\text{parallelepiped}} = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})|, \qquad V_{\text{tetrahedron}} = \frac{1}{6}|\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})|

Teen vectors coplanar hote hain agar unka scalar triple product zero ho. Similarly, chaar points coplanar hote hain agar unse bane teen vectors ka triple product zero ho.

Key point: Triple product \(= 0\) is the coplanarity test — for three vectors or for four points.

Mathematics 11 (Punjab Board)

Solved Examples: Chapter 14 – Vectors in Space

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 14: Vectors in Space, worked in full.

Example 1: Given \(\mathbf{u} = [1, -2, 3]\), \(\mathbf{v} = [2, 1, 3]\), \(\mathbf{w} = [-1, 4, 0]\), find: (i) \(\mathbf{v} + \mathbf{w}\) (ii) \(2\mathbf{w}\) (iii) \(|\mathbf{u}|\) (iv) \(|\mathbf{v} - 2\mathbf{w}|\) (v) \(|2\mathbf{u} - \mathbf{v} + 3\mathbf{w}|\)

Vector addition aur scalar multiplication component-wise hoti hai. Magnitude ke liye \(\sqrt{x^2+y^2+z^2}\) ka formula use karenge.

(i)

\mathbf{v} + \mathbf{w} = [2-1, 1+4, 3+0] = [1, 5, 3]

(ii)

2\mathbf{w} = 2[-1, 4, 0] = [-2, 8, 0]

(iii)

|\mathbf{u}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}

(iv)

\mathbf{v} - 2\mathbf{w} = [2,1,3] - [-2,8,0] = [4,-7,3], \quad |\mathbf{v}-2\mathbf{w}| = \sqrt{16+49+9} = \sqrt{74}

(v)

2\mathbf{u} - \mathbf{v} + 3\mathbf{w} = [2,-4,6] - [2,1,3] + [-3,12,0] = [-3,7,3]

|2\mathbf{u} - \mathbf{v} + 3\mathbf{w}| = \sqrt{9+49+9} = \sqrt{67}

Answer: (i) \([1, 5, 3]\), (ii) \([-2, 8, 0]\), (iii) \(\sqrt{14}\), (iv) \(\sqrt{74}\), (v) \(\sqrt{67}\)

Example 2: Find a unit vector in the direction of \(\mathbf{u} = 2\mathbf{i} + 5\mathbf{j} - \mathbf{k}\)

Unit vector ka formula hai \(\hat{\mathbf{u}} = \frac{\mathbf{u}}{|\mathbf{u}|}\). Pehle magnitude nikalenge, phir divide karenge.

|\mathbf{u}| = \sqrt{2^2 + 5^2 + (-1)^2} = \sqrt{30}

\hat{\mathbf{u}} = \frac{2\mathbf{i} + 5\mathbf{j} - \mathbf{k}}{\sqrt{30}}

Answer: \(\frac{1}{\sqrt{30}}(2\mathbf{i} + 5\mathbf{j} - \mathbf{k})\)

Example 3: Given \(\mathbf{u} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\), \(\mathbf{v} = 4\mathbf{i} + 6\mathbf{j} + 2\mathbf{k}\), \(\mathbf{w} = -6\mathbf{i} - 9\mathbf{j} - 3\mathbf{k}\), show that the three vectors are parallel to each other.

Do vectors parallel hote hain agar ek doosre ka scalar multiple ho. \(\mathbf{v}\) aur \(\mathbf{w}\) ko \(\mathbf{u}\) ke multiple ki form mein likhenge.

\mathbf{v} = 4\mathbf{i} + 6\mathbf{j} + 2\mathbf{k} = 2(2\mathbf{i} + 3\mathbf{j} + \mathbf{k}) = 2\mathbf{u}

\mathbf{w} = -6\mathbf{i} - 9\mathbf{j} - 3\mathbf{k} = -3(2\mathbf{i} + 3\mathbf{j} + \mathbf{k}) = -3\mathbf{u}

Since \(\mathbf{v} = 2\mathbf{u}\) and \(\mathbf{w} = -3\mathbf{u}\), all three are scalar multiples of \(\mathbf{u}\).

Answer: \(\mathbf{v} = 2\mathbf{u}\), \(\mathbf{w} = -3\mathbf{u}\) — hence all three are parallel.

Example 4: Find the distance between the points \((2, 4, 7)\) and \((6, 1, 3)\) in space.

3D distance formula use karenge: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\).

d = \sqrt{(6-2)^2 + (1-4)^2 + (3-7)^2} = \sqrt{16 + 9 + 16} = \sqrt{41}

Answer: \(\sqrt{41}\) units

Example 5: Prove the component formula for the dot product using the law of cosines.

Triangle ke teesre side \(\mathbf{v} - \mathbf{w}\) par law of cosines lagayenge, aur \(|\mathbf{v} - \mathbf{w}|^2\) ko components mein expand karke compare karenge.

Let \(\mathbf{v} = [x_1, y_1, z_1]\), \(\mathbf{w} = [x_2, y_2, z_2]\) with angle \(\theta\) between them. By the law of cosines:

|\mathbf{v} - \mathbf{w}|^2 = |\mathbf{v}|^2 + |\mathbf{w}|^2 - 2|\mathbf{v}||\mathbf{w}|\cos\theta

Expanding in components:

|\mathbf{v} - \mathbf{w}|^2 = (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2 = |\mathbf{v}|^2 + |\mathbf{w}|^2 - 2(x_1x_2 + y_1y_2 + z_1z_2)

Comparing the two expressions gives \(|\mathbf{v}||\mathbf{w}|\cos\theta = x_1x_2 + y_1y_2 + z_1z_2\).

Answer: \(\mathbf{v} \cdot \mathbf{w} = x_1x_2 + y_1y_2 + z_1z_2 = |\mathbf{v}||\mathbf{w}|\cos\theta\). Proved.

Example 6: If \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), prove that the projections of \(\mathbf{v}\) along \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are \(a, b, c\) respectively.

Projection ka formula hai \(\frac{\mathbf{v} \cdot \text{(unit vector)}}{|\text{unit vector}|}\). \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) unit vectors hain.

\text{Projection along } \mathbf{i} = \frac{\mathbf{v} \cdot \mathbf{i}}{|\mathbf{i}|} = (a\mathbf{i}+b\mathbf{j}+c\mathbf{k}) \cdot \mathbf{i} = a

Similarly, projection along \(\mathbf{j}\) is \(b\) and along \(\mathbf{k}\) is \(c\).

Answer: The projections are \(a\), \(b\), \(c\). Proved.

Example 7: Prove the law of cosines \(a^2 = b^2 + c^2 - 2bc\cos A\) using vectors.

Triangle ki sides par vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) lein ge jahan \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\). Phir dot product lekar expand karenge.

Let vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) lie along sides \(BC, CA, AB\) of \(\triangle ABC\), so \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\), i.e. \(\mathbf{a} = -(\mathbf{b} + \mathbf{c})\).

\mathbf{a} \cdot \mathbf{a} = (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c}) = |\mathbf{b}|^2 + |\mathbf{c}|^2 + 2\,\mathbf{b} \cdot \mathbf{c}

The angle between \(\mathbf{b}\) (along \(CA\)) and \(\mathbf{c}\) (along \(AB\)) is \(180^\circ - A\), so \(\mathbf{b} \cdot \mathbf{c} = bc\cos(180^\circ - A) = -bc\cos A\).

a^2 = b^2 + c^2 - 2bc\cos A

Answer: \(a^2 = b^2 + c^2 - 2bc\cos A\). Proved.

Example 8: Prove that \(\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta\) using vectors.

xy-plane mein do unit vectors \(\overrightarrow{OA}\) aur \(\overrightarrow{OB}\) lein ge jo positive x-axis ke sath \(\alpha\) aur \(\beta\) angles banate hain. Unka dot product do tareeqon se nikalenge.

\overrightarrow{OA} = \cos\alpha\,\mathbf{i} + \sin\alpha\,\mathbf{j}, \qquad \overrightarrow{OB} = \cos\beta\,\mathbf{i} + \sin\beta\,\mathbf{j}

The angle between them is \(\alpha - \beta\), so \(\overrightarrow{OA} \cdot \overrightarrow{OB} = (1)(1)\cos(\alpha - \beta)\). Also by components:

\overrightarrow{OA} \cdot \overrightarrow{OB} = \cos\alpha\cos\beta + \sin\alpha\sin\beta

\therefore \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta

Answer: Proved.

Example 9: Given \(\mathbf{u} = 3\mathbf{i} - \mathbf{j} - 2\mathbf{k}\) and \(\mathbf{v} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\), find \(\mathbf{u} \cdot \mathbf{v}\).

Dot product: corresponding components multiply karke add kar dete hain.

\mathbf{u} \cdot \mathbf{v} = (3)(1) + (-1)(2) + (-2)(-1) = 3 - 2 + 2 = 3

Answer: \(3\)

Example 10: Given \(\mathbf{u} = 2\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}\) and \(\mathbf{v} = 4\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\), show that \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular.

Perpendicular vectors ka dot product zero hota hai. Dot product calculate karke check karenge.

\mathbf{u} \cdot \mathbf{v} = (2)(4) + (-4)(-3) + (5)(-4) = 8 + 12 - 20 = 0

Since the dot product is zero, the vectors are perpendicular.

Answer: \(\mathbf{u} \cdot \mathbf{v} = 0\), so \(\mathbf{u} \perp \mathbf{v}\).

Example 11: \(\mathbf{u} = 2\mathbf{i} + \alpha\mathbf{j} + 5\mathbf{k}\) and \(\mathbf{v} = 3\mathbf{i} + \mathbf{j} + \alpha\mathbf{k}\) are orthogonal. Find \(\alpha\).

Orthogonal vectors ka dot product zero hota hai. \(\mathbf{u} \cdot \mathbf{v} = 0\) set karke \(\alpha\) solve karenge.

\mathbf{u} \cdot \mathbf{v} = (2)(3) + (\alpha)(1) + (5)(\alpha) = 6 + 6\alpha = 0

\Rightarrow \alpha = -1

Answer: \(\alpha = -1\)

Example 12: Find the angle between \(\mathbf{u} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\) and \(\mathbf{v} = -\mathbf{i} + \mathbf{j}\).

Angle ka formula hai \(\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\). Pehle dot product aur magnitudes nikalenge.

\mathbf{u} \cdot \mathbf{v} = (2)(-1) + (-1)(1) + (1)(0) = -3

|\mathbf{u}| = \sqrt{4+1+1} = \sqrt{6}, \qquad |\mathbf{v}| = \sqrt{1+1} = \sqrt{2}

\cos\theta = \frac{-3}{\sqrt{6}\,\sqrt{2}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2} \quad \Rightarrow \quad \theta = 150^\circ

Answer: \(150^\circ\)

Example 13: Given \(\overrightarrow{AB} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\), \(\overrightarrow{BC} = \mathbf{i} - 3\mathbf{j} - 5\mathbf{k}\) and \(\overrightarrow{AC} = 3\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}\), show that these vectors form a right triangle.

Pehle check karenge ke \(\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\) (triangle banta hai). Phir right angle ke liye kisi vertex par dot product zero hona chahiye.

\overrightarrow{AB} + \overrightarrow{BC} = (2+1)\mathbf{i} + (-1-3)\mathbf{j} + (1-5)\mathbf{k} = 3\mathbf{i} - 4\mathbf{j} - 4\mathbf{k} = \overrightarrow{AC}

So the vectors form a triangle. Now:

\overrightarrow{AB} \cdot \overrightarrow{BC} = (2)(1) + (-1)(-3) + (1)(-5) = 2 + 3 - 5 = 0

Since \(\overrightarrow{AB} \perp \overrightarrow{BC}\), the triangle is right-angled at \(B\).

Answer: The vectors form a triangle with a right angle at \(B\).

Example 14: Forces \(2\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}\) and \(-\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) act on a body displaced from \(P(4, -3, -2)\) to \(Q(6, 1, -3)\). Find the total work done.

Pehle total force add karenge, phir displacement \(\overrightarrow{PQ}\) nikalenge. Work done \(W = \mathbf{F} \cdot \mathbf{d}\).

\mathbf{F} = (2\mathbf{i}+5\mathbf{j}+6\mathbf{k}) + (-\mathbf{i}-2\mathbf{j}-\mathbf{k}) = \mathbf{i} + 3\mathbf{j} + 5\mathbf{k}

\mathbf{d} = \overrightarrow{PQ} = (6-4)\mathbf{i} + (1+3)\mathbf{j} + (-3+2)\mathbf{k} = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k}

W = \mathbf{F} \cdot \mathbf{d} = (1)(2) + (3)(4) + (5)(-1) = 2 + 12 - 5 = 9

Answer: \(9\) units of work

Example 15: Given \(\mathbf{a} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = 4\mathbf{i} + 2\mathbf{j} - \mathbf{k}\), find \(\mathbf{a} \times \mathbf{b}\) and verify that it is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

Cross product determinant method se nikalenge. Perpendicular hone ki verification ke liye dot product zero check karenge.

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 4 & 2 & -1 \end{vmatrix} = -\mathbf{i} + 6\mathbf{j} + 8\mathbf{k}

Verification:

\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = (2)(-1) + (-1)(6) + (1)(8) = 0

\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = (4)(-1) + (2)(6) + (-1)(8) = 0

Answer: \(\mathbf{a} \times \mathbf{b} = -\mathbf{i} + 6\mathbf{j} + 8\mathbf{k}\), perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).

Example 16: Given \(\mathbf{a} = 4\mathbf{i} + 3\mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\), find a unit vector perpendicular to both, and find \(\sin\theta\) where \(\theta\) is the angle between them.

Unit perpendicular vector \(= \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}\). Sine ke liye formula \(\sin\theta = \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}\) use karenge.

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & 1 \\ 2 & -1 & 2 \end{vmatrix} = 7\mathbf{i} - 6\mathbf{j} - 10\mathbf{k}

|\mathbf{a} \times \mathbf{b}| = \sqrt{49 + 36 + 100} = \sqrt{185}

\text{Unit vector} = \frac{7\mathbf{i} - 6\mathbf{j} - 10\mathbf{k}}{\sqrt{185}}

|\mathbf{a}| = \sqrt{26}, \quad |\mathbf{b}| = 3 \quad \Rightarrow \quad \sin\theta = \frac{\sqrt{185}}{3\sqrt{26}}

Answer: unit vector \(\frac{7\mathbf{i} - 6\mathbf{j} - 10\mathbf{k}}{\sqrt{185}}\); \(\sin\theta = \frac{\sqrt{185}}{3\sqrt{26}}\)

Example 17: Prove that \(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta\) using the cross product.

xy-plane mein do unit vectors \(\overrightarrow{OA}\) aur \(\overrightarrow{OB}\) lein ge jo positive x-axis ke sath \(\alpha\) aur \(-\beta\) angles banate hain. Unka cross product do tareeqon se nikalenge.

\overrightarrow{OA} = \cos\alpha\,\mathbf{i} + \sin\alpha\,\mathbf{j}, \qquad \overrightarrow{OB} = \cos\beta\,\mathbf{i} - \sin\beta\,\mathbf{j}

The angle from \(\overrightarrow{OA}\) to \(\overrightarrow{OB}\) is \(\alpha + \beta\), so \(\overrightarrow{OA} \times \overrightarrow{OB} = -\sin(\alpha+\beta)\,\mathbf{k}\) (clockwise). By components:

\overrightarrow{OA} \times \overrightarrow{OB} = -(\cos\alpha\sin\beta + \sin\alpha\cos\beta)\,\mathbf{k} = -\sin(\alpha+\beta)\,\mathbf{k}

\therefore \sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta

Answer: Proved.

Example 18: Prove the law of sines \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) using vectors.

Sides par vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) lein ge jahan \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\). Cross product lekar magnitudes compare karenge.

Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) lie along \(BC, CA, AB\) with \(\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}\). Taking the cross product with \(\mathbf{c}\):

\mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c} = \mathbf{0} \quad \Rightarrow \quad \mathbf{b} \times \mathbf{c} = \mathbf{c} \times \mathbf{a}

Taking magnitudes, and noting the angle between consecutive side-vectors is \(180^\circ\) minus the interior angle:

bc\sin A = ca\sin B \quad \Rightarrow \quad \frac{a}{\sin A} = \frac{b}{\sin B}

Similarly, crossing with \(\mathbf{a}\) gives \(\frac{b}{\sin B} = \frac{c}{\sin C}\).

Answer: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\). Proved.

Example 19: Given \(\mathbf{u} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\) and \(\mathbf{v} = 4\mathbf{i} + 2\mathbf{j} - \mathbf{k}\), find: (i) \(\mathbf{u} \times \mathbf{u}\) (ii) \(\mathbf{u} \times \mathbf{v}\) (iii) \(\mathbf{v} \times \mathbf{u}\)

Determinant method use karenge. (i) mein do rows same hon gi to answer zero. (iii) se anti-commutative property verify hogi.

(i)

\mathbf{u} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 2 & -1 & 1 \end{vmatrix} = \mathbf{0} \quad (\text{two rows identical})

(ii)

\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 4 & 2 & -1 \end{vmatrix} = (1-2)\mathbf{i} - (-2-4)\mathbf{j} + (4+4)\mathbf{k} = -\mathbf{i} + 6\mathbf{j} + 8\mathbf{k}

(iii)

\mathbf{v} \times \mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix} = (2-1)\mathbf{i} - (4+2)\mathbf{j} + (-4-4)\mathbf{k} = \mathbf{i} - 6\mathbf{j} - 8\mathbf{k}

Answer: (i) \(\mathbf{0}\), (ii) \(-\mathbf{i} + 6\mathbf{j} + 8\mathbf{k}\), (iii) \(\mathbf{i} - 6\mathbf{j} - 8\mathbf{k} = -(\mathbf{u} \times \mathbf{v})\)

Example 20: The vertices of a parallelogram are \(P(0,0,0)\), \(Q(-1,2,4)\), \(R(2,-1,4)\), \(S(1,1,8)\). Find its area.

Parallelogram ka area \(= |\overrightarrow{PQ} \times \overrightarrow{PR}|\) hota hai, jahan \(\overrightarrow{PQ}\) aur \(\overrightarrow{PR}\) adjacent sides hain.

\overrightarrow{PQ} = -\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}, \qquad \overrightarrow{PR} = 2\mathbf{i} - \mathbf{j} + 4\mathbf{k}

\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 4 \\ 2 & -1 & 4 \end{vmatrix} = 12\mathbf{i} + 12\mathbf{j} - 3\mathbf{k}

\text{Area} = \sqrt{144 + 144 + 9} = \sqrt{297} = 3\sqrt{33}

Answer: \(3\sqrt{33}\) square units

Example 21: Given \(A(1,-1,1)\), \(B(2,1,-1)\), \(C(-1,1,2)\), find the area of \(\triangle ABC\) and a unit vector perpendicular to its plane.

Triangle ka area \(= \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|\). Unit perpendicular vector \(= \frac{\overrightarrow{AB} \times \overrightarrow{AC}}{|\overrightarrow{AB} \times \overrightarrow{AC}|}\).

\overrightarrow{AB} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}, \qquad \overrightarrow{AC} = -2\mathbf{i} + 2\mathbf{j} + \mathbf{k}

\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -2 \\ -2 & 2 & 1 \end{vmatrix} = 6\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}

|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{36+9+36} = 9 \quad \Rightarrow \quad \text{Area} = \frac{9}{2}

\text{Unit vector} = \frac{6\mathbf{i}+3\mathbf{j}+6\mathbf{k}}{9} = \frac{2\mathbf{i}+\mathbf{j}+2\mathbf{k}}{3}

Answer: Area \(\frac{9}{2}\) square units; unit vector \(\frac{2\mathbf{i}+\mathbf{j}+2\mathbf{k}}{3}\)

Example 22: Find the moment about the point \(M(-2, 4, -6)\) of the force represented by \(\overrightarrow{AB}\), where \(A(1, 2, -3)\) and \(B(3, -4, 2)\).

Moment \(= \mathbf{r} \times \mathbf{F}\), jahan \(\mathbf{r} = \overrightarrow{MA}\) (point se force ke start tak) aur \(\mathbf{F} = \overrightarrow{AB}\).

\mathbf{F} = \overrightarrow{AB} = 2\mathbf{i} - 6\mathbf{j} + 5\mathbf{k}, \qquad \mathbf{r} = \overrightarrow{MA} = 3\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}

\mathbf{M} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 3 \\ 2 & -6 & 5 \end{vmatrix} = 8\mathbf{i} - 9\mathbf{j} - 14\mathbf{k}

|\mathbf{M}| = \sqrt{64 + 81 + 196} = \sqrt{341}

Answer: \(8\mathbf{i} - 9\mathbf{j} - 14\mathbf{k}\) (magnitude \(\sqrt{341}\))

Example 23: Prove that the scalar triple product \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})\) equals the determinant \(\begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}\).

Pehle \(\mathbf{v} \times \mathbf{w}\) determinant se nikalenge, phir \(\mathbf{u}\) ke sath dot product karke dekhenge ke yehi \(3 \times 3\) determinant ki expansion hai.

\mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} = (v_2w_3-v_3w_2)\mathbf{i} + (v_3w_1-v_1w_3)\mathbf{j} + (v_1w_2-v_2w_1)\mathbf{k}

Dotting with \(\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k}\):

\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = u_1(v_2w_3-v_3w_2) + u_2(v_3w_1-v_1w_3) + u_3(v_1w_2-v_2w_1)

This is exactly the expansion of the \(3 \times 3\) determinant along its first row.

Answer: \(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}\). Proved.

Example 24: Evaluate the scalar triple product \(\mathbf{i} \cdot (\mathbf{j} \times \mathbf{k})\).

Unit vectors ko components mein likh kar determinant form use karenge.

\mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1(1-0) = 1

Answer: \(1\)

Example 25: Given \(\mathbf{u} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\), \(\mathbf{v} = -\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\), \(\mathbf{w} = \mathbf{i} - 7\mathbf{j} - 4\mathbf{k}\), find the volume of the parallelepiped formed by them.

Volume \(= |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})|\) — scalar triple product ka absolute value, determinant se nikalenge.

\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} 1 & 2 & -1 \\ -1 & 2 & 3 \\ 1 & -7 & -4 \end{vmatrix}

\[= 1(-8+21) - 2(4-3) - 1(7-2) = 13 - 2 - 5 = 6\]

\[V = |6| = 6\]

Answer: \(6\) cubic units

Example 26: Given \(A(2,1,8)\), \(B(3,2,9)\), \(C(2,1,4)\), \(D(3,3,0)\), find the volume of the tetrahedron \(ABCD\).

Tetrahedron ka volume \(= \frac{1}{6}|\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD})|\). Pehle teen edge vectors nikalenge.

\overrightarrow{AB} = \mathbf{i}+\mathbf{j}+\mathbf{k}, \quad \overrightarrow{AC} = -4\mathbf{k}, \quad \overrightarrow{AD} = \mathbf{i}+2\mathbf{j}-8\mathbf{k}

\overrightarrow{AC} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & -4 \\ 1 & 2 & -8 \end{vmatrix} = 8\mathbf{i} - 4\mathbf{j}

\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = (1)(8) + (1)(-4) + (1)(0) = 4

V = \frac{1}{6}|4| = \frac{2}{3}

Answer: \(\frac{2}{3}\) cubic units

Example 27: Given \(A(-3,5,-4)\), \(B(-1,1,1)\), \(C(-1,2,2)\), \(D(-3,4,-5)\), show that the four points are coplanar.

Chaar points coplanar hote hain agar unse bane teen vectors ka scalar triple product zero ho. \(\overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}\) ka triple product nikalenge.

\overrightarrow{AB} = 2\mathbf{i}-4\mathbf{j}+5\mathbf{k}, \quad \overrightarrow{AC} = 2\mathbf{i}-3\mathbf{j}+6\mathbf{k}, \quad \overrightarrow{AD} = -\mathbf{j}-\mathbf{k}

\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = \begin{vmatrix} 2 & -4 & 5 \\ 2 & -3 & 6 \\ 0 & -1 & -1 \end{vmatrix}

\[= 2(3+6) + 4(-2-0) + 5(-2-0) = 18 - 8 - 10 = 0\]

Since the scalar triple product is zero, the vectors are coplanar.

Answer: The triple product is \(0\), so the points are coplanar.

Example 28: The vectors \(\alpha\mathbf{i} + \mathbf{j} + \mathbf{k}\), \(\mathbf{i} + \mathbf{j} + 3\mathbf{k}\) and \(2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\) are coplanar. Find \(\alpha\).

Coplanar vectors ka scalar triple product zero hota hai. Determinant zero set karke \(\alpha\) solve karenge.

\begin{vmatrix} \alpha & 1 & 1 \\ 1 & 1 & 3 \\ 2 & 1 & -2 \end{vmatrix} = 0

\alpha(-2-3) - 1(-2-6) + 1(1-2) = -5\alpha + 8 - 1 = 0

\Rightarrow \alpha = \frac{7}{5}

Answer: \(\alpha = \frac{7}{5}\)

Example 29: A plumber exerts a force of \(30\) pounds along the negative y-axis on a lever. The pivot is at the origin and the force is applied at \((1.2, 0.5, 0)\). Find the torque.

Torque \(\mathbf{\tau} = \mathbf{r} \times \mathbf{F}\), jahan \(\mathbf{r}\) position vector hai aur \(\mathbf{F} = -30\mathbf{j}\).

\mathbf{r} = 1.2\mathbf{i} + 0.5\mathbf{j}, \qquad \mathbf{F} = -30\mathbf{j}

\mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1.2 & 0.5 & 0 \\ 0 & -30 & 0 \end{vmatrix} = -36\mathbf{k}

|\mathbf{\tau}| = 36

Answer: \(-36\mathbf{k}\) pound-feet (magnitude \(36\))

Example 30: A crane exerts the force \(\mathbf{F} = [4500, 3300, 2140]\) Newton. Find the magnitude of the force.

Magnitude ka formula \(|\mathbf{F}| = \sqrt{x^2+y^2+z^2}\) use karenge.

|\mathbf{F}| = \sqrt{4500^2 + 3300^2 + 2140^2} = \sqrt{20250000 + 10890000 + 4579600}

= \sqrt{35719600} \approx 5976.59

Answer: approximately \(5976.59\) Newton

Example 31: A store sells jackets, shoes and handbags. The quantities sold are given by \(\mathbf{u} = 300\mathbf{i} + 250\mathbf{j} + 180\mathbf{k}\) and the prices per unit by \(\mathbf{v} = 3500\mathbf{i} + 4200\mathbf{j} + 6840\mathbf{k}\). Find the total revenue.

Total revenue quantity aur price vectors ka dot product hota hai: har item ki quantity × price, phir sum.

\text{Revenue} = \mathbf{u} \cdot \mathbf{v} = (300)(3500) + (250)(4200) + (180)(6840)

\[= 1050000 + 1050000 + 1231200 = 3331200\]

Answer: \(3{,}331{,}200\) rupees

Mathematics 11 (Punjab Board)

MCQs: Chapter 14 – Vectors in Space

The magnitude of \(3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}\) is
1. \(5\)
2. \(13\)
3. \(14\)
4. \(169\)
\(\sqrt{9 + 16 + 144} = \sqrt{169} = 13\). Components ke squares ka sum phir square root.
A unit vector in the direction of \(6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\) is
1. \(\frac{6\mathbf{i}-2\mathbf{j}+3\mathbf{k}}{49}\)
2. \(7(6\mathbf{i}-2\mathbf{j}+3\mathbf{k})\)
3. \(\frac{6\mathbf{i}-2\mathbf{j}+3\mathbf{k}}{7}\)
4. \(\frac{-6\mathbf{i}+2\mathbf{j}-3\mathbf{k}}{7}\)
Magnitude \(\sqrt{36+4+9} = 7\); vector ko 7 se divide karo. Sign mat badlo.
If \(\mathbf{u} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and \(\mathbf{v} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\), then \(\mathbf{u} \cdot \mathbf{v} =\)
1. \(3\)
2. \(-3\)
3. \(7\)
4. \(0\)
\((1)(2) + (2)(-1) + (-1)(3) = 2 - 2 - 3 = -3\).
If \(\mathbf{u} \cdot \mathbf{v} = 0\) for non-zero vectors \(\mathbf{u}, \mathbf{v}\), then they are
1. parallel
2. perpendicular
3. equal
4. zero vectors
Dot product zero = perpendicular hone ki condition. Yehi sab se zyada poocha jata hai.
The direction cosines of \(\mathbf{i} + \mathbf{j} + \mathbf{k}\) are
1. \(1, 1, 1\)
2. \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
3. \(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\)
4. \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\)
Magnitude \(\sqrt{3}\); har component ko \(\sqrt{3}\) se divide karo.
\((2\mathbf{i} - \mathbf{j} + \mathbf{k}) \times (\mathbf{i} + \mathbf{j} - \mathbf{k}) =\)
1. \(3\mathbf{j} + 3\mathbf{k}\)
2. \(-3\mathbf{j} - 3\mathbf{k}\)
3. \(3\mathbf{i} + 3\mathbf{k}\)
4. \(\mathbf{0}\)
Determinant: \(\mathbf{i}(1-1) - \mathbf{j}(-2-1) + \mathbf{k}(2+1) = 3\mathbf{j} + 3\mathbf{k}\).
Which triple can be the direction angles of a single vector?
1. \(45^\circ, 45^\circ, 60^\circ\)
2. \(30^\circ, 45^\circ, 60^\circ\)
3. \(45^\circ, 60^\circ, 60^\circ\)
4. \(60^\circ, 60^\circ, 60^\circ\)
\(\cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1\) sirf (C) mein poora hota hai: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{4} = 1\).
The projection of \(3\mathbf{i} + 4\mathbf{j}\) along \(\mathbf{i}\) is
1. \(5\)
2. \(4\)
3. \(25\)
4. \(3\)
\(\frac{(3\mathbf{i}+4\mathbf{j}) \cdot \mathbf{i}}{|\mathbf{i}|} = 3\). i-component hi projection hai.
If \(\mathbf{a}+\mathbf{b}+\mathbf{c} = \mathbf{0}\) with \(|\mathbf{a}| = 3, |\mathbf{b}| = 4, |\mathbf{c}| = 5\), the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is
1. \(60^\circ\)
2. \(45^\circ\)
3. \(90^\circ\)
4. \(30^\circ\)
\(9 + 16 + 24\cos\theta = 25 \Rightarrow \cos\theta = 0 \Rightarrow \theta = 90^\circ\).
The volume of the parallelepiped with coterminous edges \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) is
1. \(0\)
2. \(6\)
3. \(3\)
4. \(1\)
\(|\mathbf{i} \cdot (\mathbf{j} \times \mathbf{k})| = 1\) — unit cube ka volume.
For any vector \(\mathbf{u}\), \(\mathbf{u} \times \mathbf{u} =\)
1. \(\mathbf{u}\)
2. \(\mathbf{0}\)
3. \(|\mathbf{u}|\)
4. \(1\)
Determinant mein do rows same hon to value zero. Har vector apne aap ke parallel hai.
The area of the triangle with vertices \((0,0,0), (1,0,0), (0,1,0)\) is
1. \(1\)
2. \(\frac{1}{2}\)
3. \(2\)
4. \(\sqrt{2}\)
\(\frac{1}{2}|\mathbf{i} \times \mathbf{j}| = \frac{1}{2}|\mathbf{k}| = \frac{1}{2}\).
The work done by \(\mathbf{F} = 3\mathbf{i} + 4\mathbf{j}\) in displacing a particle from \((0,0,0)\) to \((1,1,0)\) is
1. \(7\)
2. \(25\)
3. \(5\)
4. \(1\)
\(W = \mathbf{F} \cdot \mathbf{d} = (3)(1) + (4)(1) = 7\).
If \(\mathbf{u} = 2\mathbf{v}\), then \(\mathbf{u}\) and \(\mathbf{v}\) are
1. perpendicular
2. parallel
3. orthogonal
4. antiparallel
Scalar multiple (positive) = parallel aur same direction. Yehi parallel ki definition hai.
\(\cos^2\alpha + \cos^2\beta + \cos^2\gamma =\)
1. \(0\)
2. \(3\)
3. \(-1\)
4. \(1\)
Direction cosines ki fundamental identity — yaad rakhna lazmi hai.
\(\mathbf{i} \cdot (\mathbf{j} \times \mathbf{k}) =\)
1. \(0\)
2. \(-1\)
3. \(1\)
4. \(2\)
\(\mathbf{j} \times \mathbf{k} = \mathbf{i}\), aur \(\mathbf{i} \cdot \mathbf{i} = 1\).
The distance between \((1, 2, 3)\) and \((4, 6, 3)\) is
1. \(\sqrt{41}\)
2. \(5\)
3. \(7\)
4. \(25\)
\(\sqrt{(3)^2 + (4)^2 + (0)^2} = \sqrt{25} = 5\).
If \(\mathbf{r}\) is parallel to \(\mathbf{F}\), then the moment \(\mathbf{M} = \mathbf{r} \times \mathbf{F}\) equals
1. \(\mathbf{r} \cdot \mathbf{F}\)
2. \(\mathbf{0}\)
3. \(|\mathbf{r}||\mathbf{F}|\)
4. \(\mathbf{r} + \mathbf{F}\)
Parallel vectors ka cross product zero hota hai.
If the scalar triple product of three vectors is \(12\), the volume of the tetrahedron they form is
1. \(12\)
2. \(2\)
3. \(6\)
4. \(4\)
Tetrahedron \(= \frac{1}{6} \times\) parallelepiped \(= \frac{12}{6} = 2\).
\(\sin\theta\), where \(\theta\) is the angle between \(\mathbf{i}\) and \(\mathbf{j}\), equals
1. \(0\)
2. \(\frac{1}{2}\)
3. \(1\)
4. \(\frac{\sqrt{3}}{2}\)
\(\mathbf{i} \perp \mathbf{j}\), to \(\theta = 90^\circ\) aur \(\sin 90^\circ = 1\).
If \(|\mathbf{u}| = 3, |\mathbf{v}| = 4\) and \(\mathbf{u} \cdot \mathbf{v} = 6\), the angle between them is
1. \(60^\circ\)
2. \(30^\circ\)
3. \(90^\circ\)
4. \(120^\circ\)
\(\cos\theta = \frac{6}{3 \times 4} = \frac{1}{2} \Rightarrow \theta = 60^\circ\).
\((\mathbf{a} + \mathbf{b}) \times (\mathbf{a} - \mathbf{b}) =\)
1. \(\mathbf{0}\)
2. \(2(\mathbf{a} \times \mathbf{b})\)
3. \(2(\mathbf{b} \times \mathbf{a})\)
4. \(\mathbf{a} \times \mathbf{b}\)
Expand karo: \(-\mathbf{a}\times\mathbf{b} + \mathbf{b}\times\mathbf{a} = 2(\mathbf{b}\times\mathbf{a})\). Sign ka khayal rakho.

Mathematics 11 (Punjab Board)

Chapter 14: Vectors in Space — Formula Sheet

Vectors in Space — Basics

\[\mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} + u_3\mathbf{k} = [u_1, u_2, u_3]\]

\[\mathbf{u} \pm \mathbf{v} = [u_1 \pm v_1, u_2 \pm v_2, u_3 \pm v_3], \qquad k\mathbf{u} = [ku_1, ku_2, ku_3]\]

\[\mathbf{u} \parallel \mathbf{v} \iff \mathbf{v} = c\mathbf{u} \text{ for some scalar } c\]

Magnitude, Distance, Unit Vector

\[|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}\]

\[d(P_1, P_2) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\]

\[\hat{\mathbf{u}} = \frac{\mathbf{u}}{|\mathbf{u}|}\]

Direction Angles and Direction Cosines

\[\cos\alpha = \frac{u_1}{|\mathbf{u}|}, \quad \cos\beta = \frac{u_2}{|\mathbf{u}|}, \quad \cos\gamma = \frac{u_3}{|\mathbf{u}|}\]

\[\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\]

Dot (Scalar) Product

\[\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 = |\mathbf{u}||\mathbf{v}|\cos\theta\]

\[\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}||\mathbf{v}|}, \qquad \mathbf{u} \perp \mathbf{v} \iff \mathbf{u} \cdot \mathbf{v} = 0\]

\[\text{Projection of } \mathbf{a} \text{ along } \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}\]

\[W = \mathbf{F} \cdot \mathbf{d}\]

Cross (Vector) Product

\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}, \qquad |\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin\theta\]

\[\mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u}), \qquad \mathbf{u} \times \mathbf{u} = \mathbf{0}, \qquad \mathbf{u} \times \mathbf{v} = \mathbf{0} \iff \mathbf{u} \parallel \mathbf{v}\]

\[\sin\theta = \frac{|\mathbf{u} \times \mathbf{v}|}{|\mathbf{u}||\mathbf{v}|}\]

\[|\mathbf{a} \times \mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b})^2 = |\mathbf{a}|^2|\mathbf{b}|^2\]

Area and Moment

\[A_{\text{parallelogram}} = |\mathbf{u} \times \mathbf{v}|, \qquad A_{\text{triangle}} = \frac{1}{2}|\mathbf{u} \times \mathbf{v}|\]

\[\mathbf{M} = \mathbf{r} \times \mathbf{F}\]

Scalar Triple Product

\[[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}\]

\[[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = [\mathbf{v}\;\mathbf{w}\;\mathbf{u}] = [\mathbf{w}\;\mathbf{u}\;\mathbf{v}]\]

\[V_{\text{parallelepiped}} = |[\mathbf{u}\;\mathbf{v}\;\mathbf{w}]|, \qquad V_{\text{tetrahedron}} = \frac{1}{6}|[\mathbf{u}\;\mathbf{v}\;\mathbf{w}]|\]

\[[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = 0 \iff \mathbf{u}, \mathbf{v}, \mathbf{w} \text{ are coplanar}\]

Chapter 14 FAQs

What is covered in Chapter 14 (Vectors in Space)?

This chapter covers Position vectors, direction ratios, dot product, cross product, lines and planes. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.

How many marks does Chapter 14 carry in the Punjab Board paper?

Chapter 14 (Vectors in Space) carries approximately 5 marks in the Punjab Board Class 11 Mathematics paper.

Are these notes for the new PCTB syllabus?

Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.

Where are the exercise solutions for Chapter 14?

Use the exercise index on this page — every exercise of Chapter 14 has its own page with complete step-by-step solutions.