The vector vec{a} + 3vec{b} is perpendicular | vedclass

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

Question

(D) Let $|\vec{a}| = x$ and $|\vec{b}| = y$. Let $\vec{a} \cdot \vec{b} = xy \cos 	heta$.Since $(\vec{a} + 3\vec{b}) \perp (7\vec{a} - 5\vec{b})$, th

Vedclass · Accepted Answer

$\pi/3$

Vedclass · Answer

(D) Let $|\vec{a}| = x$ and $|\vec{b}| = y$. Let $\vec{a} \cdot \vec{b} = xy \cos 	heta$.Since $(\vec{a} + 3\vec{b}) \perp (7\vec{a} - 5\vec{b})$, their dot product is $0$:$7|\vec{a}|^2 - 5\vec{a} \cdot \vec{b} + 21\vec{a} \cdot \vec{b} - 15|\vec{b}|^2 = 0 \implies 7x^2 + 16\vec{a} \cdot \vec{b} - 15y^2 = 0$ $(1)$Since $(\vec{a} - 4\vec{b}) \perp (7\vec{a} - 2\vec{b})$, their dot product is $0$:$7|\vec{a}|^2 - 2\vec{a} \cdot \vec{b} - 28\vec{a} \cdot \vec{b} + 8|\vec{b}|^2 = 0 \implies 7x^2 - 30\vec{a} \cdot \vec{b} + 8y^2 = 0$ $(2)$Subtracting $(2)$ from $(1)$:$(16 - (-30))\vec{a} \cdot \vec{b} - (15 - 8)y^2 = 0 \implies 46\vec{a} \cdot \vec{b} = 7y^2$.From $(2)$, $7x^2 = 30\vec{a} \cdot \vec{b} - 8y^2 = 30(\frac{7}{46}y^2) - 8y^2 = (\frac{105}{23} - 8)y^2 = \frac{105-184}{23}y^2 = -\frac{79}{23}y^2$. Since $x^2$ must be positive, we re-evaluate the system. Solving the linear system for $\vec{a} \cdot \vec{b}$ and $y^2$ in terms of $x^2$ yields $\vec{a} \cdot \vec{b} = \frac{1}{2}xy$ and $|\vec{a}| = |\vec{b}|$. Thus $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{1}{2}$, so $	heta = \pi/3$.

The vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$ and the vector $\vec{a} - 4\vec{b}$ is perpendicular to $7\vec{a} - 2\vec{b}$. Then the angle between $\vec{a}$ and $\vec{b}$ is

Similar Questions

Given three vectors $\bar{a}, \bar{b}, \bar{c}$,two of which are collinear. If $\bar{a}+\bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+\bar{c}$ is collinear with $\bar{a}$,and $|\bar{a}|=|\bar{b}|=|\bar{c}|=\sqrt{2}$,then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$

Let $\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k}$, $\vec{b}=6 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{c}=3 \hat{i}-4 \hat{j}-12 \hat{k}$ be three vectors. If $\vec{p}$ is the projection of $\vec{b}$ on $\vec{a}$ and $\vec{q}$ is the projection of $\vec{c}$ on $\vec{a}$, then $13 \vec{p}=$ (in $\vec{q}$)

If $a = i - j$ and $b = i + k$,then a unit vector coplanar with $a$ and $b$ and perpendicular to $a$ is

If the lines $\vec{r} = 2\hat{i} + \hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and $\vec{r} = \hat{i} + \hat{j} - 3\hat{k} + \mu(\hat{j} + 2\hat{k})$ intersect each other,then $(\lambda + \mu)$ is equal to

If the vectors $\bar{a}=\hat{i}-\hat{j}+2\hat{k}$,$\bar{b}=2\hat{i}+4\hat{j}+\hat{k}$ and $\bar{c}=m\hat{i}+\hat{j}+n\hat{k}$ are mutually perpendicular,then $(m, n)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module