Match List - I with List - II. List - I List - II (A) The value of i.…

(A) $\hat i, \hat j, \hat k$

$\hat i×\hat j=\hat k ⇒\hat j×\hat k=\hat i$,$\hat k ×\hat i=\hat j$

Do i product. $\hat i.\hat i$, $\hat j.\hat j$ = 1

$\begin{matrix}\hat i×\hat j=\hat k&\hat j × \hat i = -\hat k\\\hat j×\hat k=\hat i&\hat k×\hat i=-\hat i\\\hat k ×\hat i=\hat j&\hat i×\hat k-\hat j\end{matrix}$

$=\hat i.(\hat j ×\hat k)+\hat j.(\hat i ×\hat k)+\hat k.(\hat i×\hat j)$

$=\hat i.(\hat i)+\hat j.(-\hat j)+\hat k.(\hat k)$

$=\hat i.\hat i-\hat j.\hat j+\hat k.\hat k$

⇒ 1 - 1 + 1 = 1

(B) If $|\vec{a}|=10$, $|\vec{b}|=2$, $\vec{a}.\vec{b}=12$, then the value of $|\vec{a}×\vec{b}|$ is

$\vec a.\vec b=|\vec a||\vec b|.cosθ⇒12=10×2cosθ$

$⇒cosθ=\frac{12}{20}=\frac{3}{5}$

$⇒sinθ=\sqrt{1-cos^2θ}=\sqrt{1-\frac{9}{25}}⇒sinθ=±\frac{4}{5}$

$∴|\vec a×\vec b|=|\vec a||\vec b||sinθ|⇒10×2×\frac{4}{5}$

(C) If θ in the angle between two vectors $\vec{a}$ and $\vec{b}$, then the value of θ, for which $\vec{a}.\vec{b}=|\vec{a}×\vec{b}|$ is

value for θ $=\frac{\pi}{4}$

(D) If $\vec{a}$ and $\vec{b}$ are perpendicular and $\vec{a}=2\hat{i}+4\hat{j}+λ\hat{k}$ and $\vec{b}=3\hat{i}-5\hat{j}+\hat{k}$ then the value of λ is:

given vectors are perpendicular

$∴ (2\hat{i}+4\hat{j}+λ\hat{k}).(3\hat{i}-5\hat{j}+\hat{k})$

$(2)(3)+(4)(5)+(λ)(1)$

$6+(-20)+λ⇒6-20+λ⇒14=λ$

So, correct option is 1. -i.e., (A) - (IV), (B) - (I), (C) - (II), (D) - (III)