Question

If \( \vec{a} = \hat{i}+\hat{j}+\hat{k} \), \( \vec{b} = \hat{i}-\hat{j}+\hat{k} \) and \( \vec{c} = \hat{i}+\hat{j}-\hat{k} \), then match the following: \[ \begin{array}{|c|c|c|} \hline \text{List-I} & & \text{List-II} \\ \hline A & [\vec{a}\ \vec{b}\ \vec{c}] & I: 4 \\ \hline B & |\vec{a}+\vec{b}+\vec{c}|^2 & II: 11 \\ \hline C & \text{Volume of tetrahedron} & III: \dfrac{2}{3} \\ \hline D & |(\vec{a}\times\vec{b})\times(\vec{a}\times\vec{c})| & IV: 4\sqrt{3} \\ \hline & & V: 12 \\ \hline \end{array} \]

• A. A-I, B-II, C-III, D-V
• B. A-I, B-III, C-II, D-V
• C. A-I, B-II, C-V, D-IV
• D. A-I, B-II, C-III, D-IV

Hint:

Important identities in vector algebra: \[ [\vec{a}\ \vec{b}\ \vec{c}] = \det \begin{bmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{bmatrix} \] and \[ \text{Volume of tetrahedron} = \frac{1}{6}\times \text{Scalar Triple Product} \]

Answer 1

The Correct Option is D

Concept: This problem uses scalar triple product, vector magnitude and cross product identities. Important formulas: \[ [\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) \] \[ \text{Volume of tetrahedron} = \frac{1}{6}|[\vec{a}\ \vec{b}\ \vec{c}]| \]

Step 1: Find scalar triple product. \[ \vec{a}=(1,1,1),\quad \vec{b}=(1,-1,1),\quad \vec{c}=(1,1,-1) \] Now, \[ [\vec{a}\ \vec{b}\ \vec{c}] = \begin{vmatrix} 1&1&1\\ 1&-1&1\\ 1&1&-1 \end{vmatrix} \] Expanding determinant: \[ = 1\begin{vmatrix}-1&1\\1&-1\end{vmatrix} - 1\begin{vmatrix}1&1\\1&-1\end{vmatrix} + 1\begin{vmatrix}1&-1\\1&1\end{vmatrix} \] \[ = 1(1-1)-1(-1-1)+1(1+1) \] \[ =0+2+2 \] \[ =4 \] Thus, \[ A \rightarrow I \]

Step 2: Find \( |\vec{a}+\vec{b}+\vec{c}|^2 \). \[ \vec{a}+\vec{b}+\vec{c} = (1+1+1,\ 1-1+1,\ 1+1-1) \] \[ =(3,1,1) \] Hence, \[ |\vec{a}+\vec{b}+\vec{c}|^2 = 3^2+1^2+1^2 \] \[ =9+1+1 \] \[ =11 \] Therefore, \[ B \rightarrow II \]

Step 3: Find volume of tetrahedron. \[ \text{Volume} = \frac{1}{6}\times 4 = \frac{2}{3} \] Hence, \[ C \rightarrow III \]

Step 4: Evaluate cross product expression. After computing carefully, \[ |(\vec{a}\times\vec{b})\times(\vec{a}\times\vec{c})| = 4\sqrt{3} \] Thus, \[ D \rightarrow IV \] Hence correct matching is: \[ A-I,\ B-II,\ C-III,\ D-IV \]