Two adjacent sides of a parallelogram ABCD ar | vedclass

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

Question

(B) Step $1$: Calculate the magnitudes of $\vec{AB}$ and $\vec{AD}$.$|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.$

Vedclass · Accepted Answer

$\sqrt{17}/9$

Vedclass · Answer

(B) Step $1$: Calculate the magnitudes of $\vec{AB}$ and $\vec{AD}$.$|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.$|\vec{AD}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Step $2$: Find the angle $	heta$ between $\vec{AB}$ and $\vec{AD}$.$\cos 	heta = \frac{\vec{AB} \cdot \vec{AD}}{|\vec{AB}| |\vec{AD}|} = \frac{(2)(-1) + (10)(2) + (11)(2)}{15 	imes 3} = \frac{-2 + 20 + 22}{45} = \frac{40}{45} = \frac{8}{9}$.Step $3$: Since $\vec{AD'}$ is in the same plane and perpendicular to $\vec{AB}$, the angle between $\vec{AD'}$ and $\vec{AB}$ is $90^\circ$. The angle between $\vec{AD}$ and $\vec{AD'}$ is $\alpha$. Thus, $	heta = 90^\circ - \alpha$ (or $\alpha = 90^\circ - 	heta$).Step $4$: Calculate $\cos \alpha = \cos(90^\circ - 	heta) = \sin 	heta$.Since $\cos 	heta = 8/9$, $\sin 	heta = \sqrt{1 - (8/9)^2} = \sqrt{1 - 64/81} = \sqrt{17/81} = \sqrt{17}/9$.

Similar Questions

If $l\vec{a} + m\vec{b} + n\vec{c} = \vec{0},$ where $l, m, n$ are scalars and $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular non-zero vectors,then

If $\vec{a}=\frac{3}{2} \hat{k}$ and $\vec{b}=\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{2}$,then the angle between $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is: (in $^{\circ}$)

$A$ vector coplanar with vectors $i + j$ and $j + k$ and parallel to the vector $2i - 2j - 4k$ is:

If $A=(0,4,-3)$,$B=(5,0,12)$,and $C=(7,24,0)$,then $\angle BAC=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module