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 vectors $\vec{AB}$ and $\vec{AD}$.$|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225}

Vedclass · Accepted Answer

$\sqrt{17}/9$

Vedclass · Answer

(B) Step $1$: Calculate the magnitudes of vectors $\vec{AB}$ 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 $\vec{AD}$.$\vec{AB} \cdot \vec{AD} = (2)(-1) + (10)(2) + (11)(2) = -2 + 20 + 22 = 40$.$\cos 	heta = \frac{\vec{AB} \cdot \vec{AD}}{|\vec{AB}| |\vec{AD}|} = \frac{40}{15 	imes 3} = \frac{40}{45} = \frac{8}{9}$.Step $3$: Since $\vec{AD'}$ is obtained by rotating $\vec{AD}$ in the plane of the parallelogram such that $\vec{AD'} \perp \vec{AB}$, the angle between $\vec{AD'}$ and $\vec{AB}$ is $90^\circ$.Step $4$: The angle $\alpha$ is the angle of rotation, which is the difference between the original angle $	heta$ and the new angle $90^\circ$. Thus, $\alpha = |	heta - 90^\circ|$.$\cos \alpha = \cos(	heta - 90^\circ) = \sin 	heta$.Step $5$: Since $\cos 	heta = 8/9$, $\sin 	heta = \sqrt{1 - (8/9)^2} = \sqrt{1 - 64/81} = \sqrt{17/81} = \sqrt{17}/9$.

Similar Questions

$A(\vec{a}), B(\vec{b}), C(\vec{c}), D(\vec{d})$ are four concyclic points, such that $x \vec{a}+y \vec{b}+z \vec{c}+t \vec{d}=\vec{0}$ and $x+y+z+t=0$, where $x, y, z, t$ are constants not all zero. If the chords $AB$ and $CD$ intersect at $P$, then:

If $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$ and $|\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}|$, then $\theta$ is equal to

If $\overline{a}, \overline{b}, \overline{c}$ are unit vectors and $\theta$ is the angle between $\overline{a}$ and $\overline{c}$ and $\overline{a}+2 \overline{b}+2 \overline{c}=\overline{0}$,then $|\overline{a} \times \overline{c}|=$

Let $a, b, c$ be three vectors such that $a \neq 0$,$a \times b = 2a \times c$,$|a| = |c| = 1$,$|b| = 4$,and $|b \times c| = \sqrt{15}$. If $b - 2c = \lambda a$,then $\lambda$ equals:

Let $\vec a, \vec b, \vec c$ be three vectors such that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$. If $|\vec a| = 1, |\vec b| = 2, |\vec c| = 3$,then $|\vec a + \vec b + \vec c| = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module