If vec{a}, vec{b}, vec{c} are 3 vectors such | vedclass

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

Question

(A) Given $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,we can write $\vec{a}+\vec{b}=-\vec{c}$.Squaring both sides,we get $|\vec{a}+\vec{b}|^2 = |-\ve

Vedclass · Accepted Answer

$\cos ^{-1} \frac{2}{5}$

Vedclass · Answer

(A) Given $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,we can write $\vec{a}+\vec{b}=-\vec{c}$.Squaring both sides,we get $|\vec{a}+\vec{b}|^2 = |-\vec{c}|^2$.$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.Substituting the given magnitudes: $5^2 + 8^2 + 2|\vec{a}||\vec{b}| \cos 	heta = 11^2$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.$25 + 64 + 2(5)(8) \cos 	heta = 121$.$89 + 80 \cos 	heta = 121$.$80 \cos 	heta = 121 - 89 = 32$.$\cos 	heta = \frac{32}{80} = \frac{2}{5}$.Therefore,$	heta = \cos ^{-1} \left(\frac{2}{5}ight)$.

If $\vec{a}, \vec{b}, \vec{c}$ are $3$ vectors such that $|\vec{a}|=5, |\vec{b}|=8, |\vec{c}|=11$ and $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

Similar Questions

If $P$ and $Q$ are two points on the curve $y = 2^{x+2}$ in the rectangular Cartesian coordinate system such that $\overline{OP} \cdot \hat{i} = -1$ and $\overline{OQ} \cdot \hat{i} = 2$,then $\overline{OQ} - 4\overline{OP} = $

Let $D, E, F$ be the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$ respectively. Then $\vec{AD} + \vec{BE} + \vec{CF} = \dots$

Find the vector joining the points $P(2, 3, 0)$ and $Q(-1, -2, -4)$ directed from $P$ to $Q$.

If $\vec{a}=(p, -2, 5)$ and $\vec{b}=(1, q, -3)$ are collinear vectors,then:

If $ABCD$ is a parallelogram,$\overrightarrow{AB} = 2i + 4j - 5k$ and $\overrightarrow{AD} = i + 2j + 3k,$ then the unit vector in the direction of $\overrightarrow{BD}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module