Question

If $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2\hat{j}+\hat{k}$ then Match List-I with List-II:}

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

Hint:

Always prefer identity $|\vec{a}\times\vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2$ instead of determinant expansion.

Answer 1

The Correct Option is D

Concept: For vector operations, we use component-wise addition/subtraction, dot product formula, and magnitude relations: \[ |\vec{v}|=\sqrt{x^2+y^2+z^2}, \quad \vec{a}\cdot\vec{b}=a_1b_1+a_2b_2+a_3b_3 \] \[ |\vec{a}\times\vec{b}|=\sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2} \]

Step 1: {Write vectors in component form.}
\[ \vec{a}=(1,1,-1), \quad \vec{b}=(1,-2,1) \]

Step 2: {Find $\vec{a}+\vec{b}$.}
\[ \vec{a}+\vec{b}=(1+1,\;1-2,\;-1+1)=(2,-1,0) \] \[ |\vec{a}+\vec{b}|=\sqrt{2^2+(-1)^2+0^2}=\sqrt{5} \Rightarrow (II) \]

Step 3: {Find $\vec{a}-\vec{b}$.}
\[ \vec{a}-\vec{b}=(1-1,\;1-(-2),\;-1-1)=(0,3,-2) \] \[ |\vec{a}-\vec{b}|=\sqrt{0^2+3^2+(-2)^2}=\sqrt{13} \Rightarrow (IV) \]

Step 4: {Find dot product.}
\[ \vec{a}\cdot\vec{b}=1\cdot1+1\cdot(-2)+(-1)\cdot1 \] \[ =1-2-1=-2 \Rightarrow |\vec{a}\cdot\vec{b}|=2 \Rightarrow (I) \]

Step 5: {Find cross product magnitude.}
\[ |\vec{a}\times\vec{b}|=\sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2} \] \[ |\vec{a}|^2=1^2+1^2+(-1)^2=3,\quad |\vec{b}|^2=1^2+(-2)^2+1^2=6 \] \[ (\vec{a}\cdot\vec{b})^2=4 \] \[ |\vec{a}\times\vec{b}|=\sqrt{18-4}=\sqrt{14} \Rightarrow (III) \] Final Matching: \[ (A)-(II),\ (B)-(IV),\ (C)-(I),\ (D)-(III) \]