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Punjab Board Class 11 Mathematics

Exercise 14.4 Solutions — Vectors in Space

Complete worked solutions for Exercise 14.4 from Chapter 14 (Vectors in Space), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 14.4 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 14: Vectors in Space (Scalar Triple Product).

Question 1: Evaluate the scalar triple product \([\mathbf{u}\;\mathbf{v}\;\mathbf{w}]\): (i) \(\mathbf{u} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\), \(\mathbf{v} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}\), \(\mathbf{w} = \mathbf{i} + \mathbf{j} + 2\mathbf{k}\) (ii) \(\mathbf{u} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\), \(\mathbf{v} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\), \(\mathbf{w} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}\)

Scalar triple product determinant form mein nikalenge: pehli row se expand karke carefully signs ke sath solve karenge.

(i)

\[[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 1 & 1 & 2 \end{vmatrix} = 1(-2-3) - 2(4-3) - 1(2+1) = -5 - 2 - 3 = -10\]

(ii)

\[[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & 1 & 1 \end{vmatrix} = 2(2+3) + 1(1+9) + 1(1-6) = 10 + 10 - 5 = 15\]
Answer: (i) \(-10\), (ii) \(15\)

Question 2: Prove the cyclic property of the scalar triple product: \([\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = [\mathbf{v}\;\mathbf{w}\;\mathbf{u}] = [\mathbf{w}\;\mathbf{u}\;\mathbf{v}]\).

Determinant mein rows ki cyclic permutation do row-swaps ke barabar hoti hai, aur har swap sign flip karta hai — do swaps se sign wapis aa jata hai.

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

Moving the first row of \([\mathbf{u}\;\mathbf{v}\;\mathbf{w}]\) to the bottom needs two adjacent row-swaps, each changing the sign: \((-1)^2 = +1\). Hence the value is unchanged, and similarly for \([\mathbf{w}\;\mathbf{u}\;\mathbf{v}]\).

Answer: \([\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = [\mathbf{v}\;\mathbf{w}\;\mathbf{u}] = [\mathbf{w}\;\mathbf{u}\;\mathbf{v}]\). Proved.

Question 3: Show that the scalar triple product is zero if any two of the three vectors are equal.

Determinant ki property: agar do rows identical hon to determinant zero hota hai.

Suppose \(\mathbf{v} = \mathbf{u}\). Then:

\[[\mathbf{u}\;\mathbf{u}\;\mathbf{w}] = \begin{vmatrix} u_1 & u_2 & u_3 \\ u_1 & u_2 & u_3 \\ w_1 & w_2 & w_3 \end{vmatrix} = 0\]

because two rows are identical. The same holds if any other pair is equal.

Answer: Proved.

Question 4: Find the volume of the parallelepiped with coterminous edges \(\mathbf{u} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\), \(\mathbf{v} = \mathbf{i} + 3\mathbf{j} + \mathbf{k}\), \(\mathbf{w} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\).

Volume \(= |[\mathbf{u}\;\mathbf{v}\;\mathbf{w}]|\) — scalar triple product ka absolute value.

\[[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = \begin{vmatrix} 2 & 1 & -2 \\ 1 & 3 & 1 \\ 3 & -1 & 2 \end{vmatrix} = 2(6+1) - 1(2-3) - 2(-1-9) = 14 + 1 + 20 = 35\]
\[V = |35| = 35\]
Answer: \(35\) cubic units

Question 5: Find the volume of the tetrahedron with vertices \(A(1, 0, 2)\), \(B(3, 1, 0)\), \(C(2, -1, 3)\), \(D(0, 2, 1)\).

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

\[\overrightarrow{AB} = (2,1,-2), \quad \overrightarrow{AC} = (1,-1,1), \quad \overrightarrow{AD} = (-1,2,-1)\]
\[\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = \begin{vmatrix} 2 & 1 & -2 \\ 1 & -1 & 1 \\ -1 & 2 & -1 \end{vmatrix} = 2(1-2) - 1(-1+1) - 2(2-1) = -4\]
\[V = \frac{1}{6}|-4| = \frac{2}{3}\]
Answer: \(\frac{2}{3}\) cubic units

Question 6: Show that the points \(A(1, 1, 1)\), \(B(2, 3, 4)\), \(C(3, 2, 1)\), \(D(2, 2, 2)\) are coplanar.

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

\[\overrightarrow{AB} = (1,2,3), \quad \overrightarrow{AC} = (2,1,0), \quad \overrightarrow{AD} = (1,1,1)\]
\[\begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 0 \\ 1 & 1 & 1 \end{vmatrix} = 1(1-0) - 2(2-0) + 3(2-1) = 1 - 4 + 3 = 0\]

Since the scalar triple product is zero, the vectors (and hence the points) are coplanar.

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

Question 7: The vectors \(k\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), \(\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) and \(2\mathbf{i} + \mathbf{j} + \mathbf{k}\) are coplanar. Find \(k\).

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

\[\begin{vmatrix} k & 2 & 1 \\ 1 & 1 & 2 \\ 2 & 1 & 1 \end{vmatrix} = k(1-2) - 2(1-4) + 1(1-2) = -k + 6 - 1 = 5 - k = 0\]
\[\Rightarrow k = 5\]
Answer: \(k = 5\)

Question 8: Prove that \([\mathbf{a}+\mathbf{b}\;\;\mathbf{b}+\mathbf{c}\;\;\mathbf{c}+\mathbf{a}] = 2[\mathbf{a}\;\mathbf{b}\;\mathbf{c}]\).

Triple product rows mein linear hota hai. Expand karenge; jin terms mein vector repeat ho woh zero ho jayenge, aur baqi cyclic property se combine honge.

\[[\mathbf{a}+\mathbf{b}\;\; \mathbf{b}+\mathbf{c}\;\; \mathbf{c}+\mathbf{a}] = [\mathbf{a}\;\mathbf{b}\;\mathbf{c}] + [\mathbf{a}\;\mathbf{b}\;\mathbf{a}] + [\mathbf{a}\;\mathbf{c}\;\mathbf{c}] + [\mathbf{a}\;\mathbf{c}\;\mathbf{a}] + [\mathbf{b}\;\mathbf{b}\;\mathbf{c}] + [\mathbf{b}\;\mathbf{b}\;\mathbf{a}] + [\mathbf{b}\;\mathbf{c}\;\mathbf{c}] + [\mathbf{b}\;\mathbf{c}\;\mathbf{a}]\]

All terms with a repeated vector vanish (Question 3), leaving:

\[= [\mathbf{a}\;\mathbf{b}\;\mathbf{c}] + [\mathbf{b}\;\mathbf{c}\;\mathbf{a}] = [\mathbf{a}\;\mathbf{b}\;\mathbf{c}] + [\mathbf{a}\;\mathbf{b}\;\mathbf{c}] = 2[\mathbf{a}\;\mathbf{b}\;\mathbf{c}]\]
Answer: Proved.

Question 9: If \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) are coplanar, show that \([\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = 0\).

Coplanar hone ka matlab hai ke ek vector baqi do ka linear combination hai. Usay substitute karke triple product expand karenge.

Since \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) are coplanar, write \(\mathbf{w} = \alpha\mathbf{u} + \beta\mathbf{v}\) for some scalars \(\alpha, \beta\). Then:

\[[\mathbf{u}\;\mathbf{v}\;\mathbf{w}] = [\mathbf{u}\;\mathbf{v}\;\;\alpha\mathbf{u}+\beta\mathbf{v}] = \alpha[\mathbf{u}\;\mathbf{v}\;\mathbf{u}] + \beta[\mathbf{u}\;\mathbf{v}\;\mathbf{v}] = 0 + 0 = 0\]
Answer: Proved.

Question 10: Find the volume of the parallelepiped determined by \(\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}\), \(\mathbf{b} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\), \(\mathbf{c} = 2\mathbf{i} + 5\mathbf{j} + \mathbf{k}\).

Volume \(= |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|\) — determinant form use karenge.

\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} 3 & 1 & -2 \\ 1 & 2 & -2 \\ 2 & 5 & 1 \end{vmatrix} = 3(2+10) - 1(1+4) - 2(5-4) = 36 - 5 - 2 = 29\]
\[V = |29| = 29\]
Answer: \(29\) cubic units