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

Step 1: Write the vectors.
Take $\vec a=(1,1,1)$, $\vec b=(1,-1,1)$, $\vec c=(1,1,-1)$. We match each List-I item to its value.
Step 2: Scalar triple product (A).
The determinant of the three rows is $1(1-1)-1(-1-1)+1(1+1)=0+2+2=4$. So A pairs with I (value $4$).
Step 3: Magnitude squared of the sum (B).
$\vec a+\vec b+\vec c=(3,1,1)$, so its square length is $9+1+1=11$. So B pairs with II.
Step 4: Volume of tetrahedron (C).
Volume $=\dfrac16\,|[\vec a\ \vec b\ \vec c]|=\dfrac16\times4=\dfrac23$. So C pairs with III.
Step 5: The cross product term (D).
$(\vec a\times\vec b)\times(\vec a\times\vec c)$ uses the box product; computing it gives length $4\sqrt3$. So D pairs with IV.
Step 6: Read the matching.
Hence A-I, B-II, C-III, D-IV, which is option 4. \[ \boxed{A\text{-}I,\ B\text{-}II,\ C\text{-}III,\ D\text{-}IV} \]