overrightarrow A = 2hat i + 4hat j + 4hat k a | vedclass

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

Question

(c) $\cos 	heta = \frac{{\overrightarrow {m A} \,.\,\overrightarrow B }}{{|\overrightarrow {m A} \,|\,.\,|\,\overrightarrow B |}} = \frac{{{a_1}{

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(c) $\cos 	heta = \frac{{\overrightarrow {m A} \,.\,\overrightarrow B }}{{|\overrightarrow {m A} \,|\,.\,|\,\overrightarrow B |}} = \frac{{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}}{{|\overrightarrow {m A} \,|\,.\,|\overrightarrow B |\,}}$

$ = \frac{{2 	imes 4 + 4 	imes 2 - 4 	imes 4}}{{|\overrightarrow A |\,.\,|\overrightarrow B |}} = 0$

$	herefore 	heta = {\cos ^{ - 1}}(0^\circ )$ $ \Rightarrow \,\,\,	heta = 90^\circ $

$\overrightarrow A = 2\hat i + 4\hat j + 4\hat k$ and $\overrightarrow B = 4\hat i + 2\hat j - 4\hat k$ are two vectors. The angle between them will be ........ $^o$

Similar Questions

The angle between two vectors given by $6\hat i + 6\hat j - 3\hat k$ and $7\hat i + 4\hat j + 4\hat k$ is

The angle between vectors $(\overrightarrow {\rm{A}} \times \overrightarrow {\rm{B}} )$ and $(\overrightarrow {\rm{B}} \times \overrightarrow {\rm{A}} )$ is

Projection of vector $\vec A$ on $\vec B$ is

When $\vec A.\vec B = - |A||B|,$ then

Dot product of two mutual perpendicular vector is