The points O, A, B, C, D are such that overri | vedclass

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

Question

(D) We have $\overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} = (a - 2b) - b = a - 3b$. We have $\overrightarrow{AC} = \overrightarrow{

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) We have $\overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} = (a - 2b) - b = a - 3b$. We have $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (2a + 3b) - a = a + 3b$. Let $	heta$ be the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$. Then $\cos 	heta = \frac{\overrightarrow{BD} \cdot \overrightarrow{AC}}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = \frac{(a - 3b) \cdot (a + 3b)}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = \frac{|a|^2 - 9|b|^2}{|\overrightarrow{BD}| |\overrightarrow{AC}|}$. Given $|a| = 3|b|$,we have $|a|^2 = 9|b|^2$. Substituting this into the expression,we get $\cos 	heta = \frac{9|b|^2 - 9|b|^2}{|\overrightarrow{BD}| |\overrightarrow{AC}|} = 0$. Since $\cos 	heta = 0$,the angle $	heta = \frac{\pi}{2}$. Therefore,the correct option is $(d)$.

The points $O, A, B, C, D$ are such that $\overrightarrow{OA} = a, \overrightarrow{OB} = b, \overrightarrow{OC} = 2a + 3b$ and $\overrightarrow{OD} = a - 2b$. If $|a| = 3|b|$,then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is

Similar Questions

If $a$ is a unit vector, then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$

The projection of vector $\vec{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is:

Let $\vec{a}, \vec{b}, \vec{c}$ be $3$ vectors such that $|\vec{a}|=3, |\vec{b}|=2\sqrt{2}, |\vec{c}|=5$ and $\vec{c}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$. If the angle between the vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$, then $|\vec{a}+\vec{b}+\vec{c}|=$

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then :-

Let $\bar{u}, \bar{v}, \bar{w}$ be vectors such that $|\bar{u}|=1, |\bar{v}|=2, |\bar{w}|=3$. If the projection of $\bar{v}$ on $\bar{u}$ is equal to that of $\bar{w}$ on $\bar{u}$,and the vectors $\bar{v}, \bar{w}$ are perpendicular to each other,then $|\bar{u}-\bar{v}+\bar{w}|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module