|vec{a} times hat{i}|^2 + |vec{a} times hat{j | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.Then,$|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$.Now,$\vec{a} 	imes \hat{i} = (a_1\hat{i} + a_2\ha

Vedclass · Accepted Answer

$2|\vec{a}|^2$

Vedclass · Answer

(B) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.Then,$|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$.Now,$\vec{a} 	imes \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) 	imes \hat{i} = a_1(\hat{i} 	imes \hat{i}) + a_2(\hat{j} 	imes \hat{i}) + a_3(\hat{k} 	imes \hat{i}) = 0 - a_2\hat{k} + a_3\hat{j}$.So,$|\vec{a} 	imes \hat{i}|^2 = |a_3\hat{j} - a_2\hat{k}|^2 = a_3^2 + a_2^2$.Similarly,$|\vec{a} 	imes \hat{j}|^2 = |a_1\hat{k} - a_3\hat{i}|^2 = a_1^2 + a_3^2$.And,$|\vec{a} 	imes \hat{k}|^2 = |a_2\hat{i} - a_1\hat{j}|^2 = a_2^2 + a_1^2$.Adding these three,we get:$|\vec{a} 	imes \hat{i}|^2 + |\vec{a} 	imes \hat{j}|^2 + |\vec{a} 	imes \hat{k}|^2 = (a_3^2 + a_2^2) + (a_1^2 + a_3^2) + (a_2^2 + a_1^2) = 2(a_1^2 + a_2^2 + a_3^2) = 2|\vec{a}|^2$.

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

Similar Questions

If $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$ are two vectors,then $(\bar{a} + 2\bar{b}) \times (3\bar{a} - \bar{b}) = $

$A$ unit vector perpendicular to each of the vectors $2i - j + k$ and $3i + 4j - k$ is equal to

If $u = a - b$ and $v = a + b$ and $|a| = |b| = 2$,then $|u \times v|$ is equal to

If $a=\hat{i}+\hat{j}+\hat{k}$,$b=\hat{i}+\hat{j}+2\hat{k}$ and $c=2\hat{i}+3\hat{j}+4\hat{k}$,then the magnitude of the projection on $c$ of a unit vector that is perpendicular to both $a$ and $b$ is

If the vectors $a$ and $b$ are mutually perpendicular,then $a \times \{ a \times \{ a \times (a \times b)\} \}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module