|a times b|^2 + (a cdot b)^2 = ? | vedclass

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

Question

(B) We know that the magnitude of the cross product of two vectors $a$ and $b$ is given by $|a 	imes b| = |a| |b| \sin 	heta$,where $	heta$ is the

Vedclass · Accepted Answer

$|a|^2 |b|^2$

Vedclass · Answer

(B) We know that the magnitude of the cross product of two vectors $a$ and $b$ is given by $|a 	imes b| = |a| |b| \sin 	heta$,where $	heta$ is the angle between them.Squaring both sides,we get $|a 	imes b|^2 = |a|^2 |b|^2 \sin^2 	heta$.We also know that the dot product is given by $a \cdot b = |a| |b| \cos 	heta$.Squaring both sides,we get $(a \cdot b)^2 = |a|^2 |b|^2 \cos^2 	heta$.Adding these two expressions:$|a 	imes b|^2 + (a \cdot b)^2 = |a|^2 |b|^2 \sin^2 	heta + |a|^2 |b|^2 \cos^2 	heta$$= |a|^2 |b|^2 (\sin^2 	heta + \cos^2 	heta)$Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:$|a 	imes b|^2 + (a \cdot b)^2 = |a|^2 |b|^2$.Since $|a|^2 = a \cdot a$ and $|b|^2 = b \cdot b$,the result is $(a \cdot a)(b \cdot b)$.

$|a \times b|^2 + (a \cdot b)^2 = ?$

Similar Questions

The set of all $\alpha$,for which the vectors $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ is:

Let $a, b$ and $c$ be vectors with magnitudes $3, 4$ and $5$ respectively and $a + b + c = 0$. Then the value of $a \cdot b + b \cdot c + c \cdot a$ is:

The projection of $\bar{a} = \hat{i} - 2\hat{j} + \hat{k}$ on $\bar{b} = 2\hat{i} - \hat{j} + \hat{k}$ is

If $a, b, c$ are coplanar vectors,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module