Match List - I with List - II.

List - I	List - II
(A) The value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$	(I) 16
(B) If $\lvert\vec{a}\rvert=10$, $\lvert\vec{b}\rvert=2$ and $\vec{a}\cdot\vec{b}=12$, then the value of $\lvert\vec{a}\times\vec{b}\rvert$ is	(II) $\frac{\pi}{4}$
(C) If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then the value of $\theta$, for which $\vec{a}\cdot\vec{b} = \lvert\vec{a}\times\vec{b}\rvert$ is	(III) 14
(D) If $\vec{a}$ and $\vec{b}$ are perpendicular and $\vec{a} = 2\hat{i}+4\hat{j}+\lambda\hat{k}$ and $\vec{b} = 3\hat{i}-5\hat{j}+\hat{k}$, then the value of $\lambda$ is	(IV) 1

Choose the correct answer from the options given below :

(A) Evaluate $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$

• $\hat{j}\times\hat{k}=\hat{i}\ \Rightarrow\ \hat{i}\cdot\hat{i}=1$
• $\hat{i}\times\hat{k}=-\hat{j}\ \Rightarrow\ \hat{j}\cdot(-\hat{j})=-1$
• $\hat{i}\times\hat{j}=\hat{k}\ \Rightarrow\ \hat{k}\cdot\hat{k}=1$

Sum $= 1 - 1 + 1 = 1$. So (A) -> (IV).

(B) Use the identity $\lvert\vec{a}\rvert^{2}\lvert\vec{b}\rvert^{2} = (\vec{a}\cdot\vec{b})^{2} + \lvert\vec{a}\times\vec{b}\rvert^{2}$

$(10)^{2}(2)^{2} = (12)^{2} + \lvert\vec{a}\times\vec{b}\rvert^{2}$

$400 = 144 + \lvert\vec{a}\times\vec{b}\rvert^{2}$

$\lvert\vec{a}\times\vec{b}\rvert^{2} = 256\ \Rightarrow\ \lvert\vec{a}\times\vec{b}\rvert = 16$

So (B) -> (I).

(C) $\vec{a}\cdot\vec{b} = \lvert\vec{a}\times\vec{b}\rvert$

$\lvert\vec{a}\rvert\lvert\vec{b}\rvert\cos\theta = \lvert\vec{a}\rvert\lvert\vec{b}\rvert\sin\theta\ \Rightarrow\ \tan\theta = 1\ \Rightarrow\ \theta = \dfrac{\pi}{4}$

So (C) -> (II).

(D) Perpendicular means $\vec{a}\cdot\vec{b}=0$

$(2)(3) + (4)(-5) + (\lambda)(1) = 0$

$6 - 20 + \lambda = 0\ \Rightarrow\ \lambda = 14$

So (D) -> (III).