Match List I with List II

Choose the correct answer from the options given below :

A-IV, B-I,C-II, D-III

The correct answer is Option (4) → A-IV, B-I,C-II, D-III

(A) $\vec a=a_x\hat i+a_y\hat j+a_z\hat k$

$|\vec{a}×\hat{i}|^2+|\vec{a}×\hat{j}|^2+|\vec{a}×\hat{k}|^2$

$=|-a_y\hat k+a_z\hat j|^2+|a_x\hat k-a_z\hat i|^2+|-a_x\hat j+a_y\hat i|^2$

$={a_y}^2+{a_z}^2+{a_x}^2+{a_z}^2+{a_x}^2+{a_y}^2$

$=2|\vec a|^2$ (IV)

(B) $\frac{(\vec{a}.\vec{b})^2+|\vec{a}×\vec{b}|^2}{|\vec{b}|^2}=\frac{|\vec{a}|^2|\vec{b}|^2(\cos^2θ+\sin^θ)}{|\vec{b}|^2}=|\vec{a}|^2$ (I)

(C) let $\vec a=a\hat i⇒b=b\hat j⇒c=c\hat k$

$a+b+c$

so $|\vec a+\vec b+\vec c|=|a\hat i+b\hat j+c\hat k|$

$=\sqrt{3}a=\sqrt{3}|\vec a|$ (II)

(D) area = $\frac{1}{2}|\vec{a}×\vec{b}|$ (III)