The vector vec{a} = (alpha, 2, beta) lies in | vedclass

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

Question

(D) Since $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar,their scalar triple product is zero:$[\vec{a} \vec{b} \vec{c}] = 0$$\begin{vmatrix} \alpha &

Vedclass · Accepted Answer

$\alpha = 1, \beta = 1$

Vedclass · Answer

(D) Since $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar,their scalar triple product is zero:$[\vec{a} \vec{b} \vec{c}] = 0$$\begin{vmatrix} \alpha & 2 & \beta \ 1 & 1 & 0 \ 0 & 1 & 1 \end{vmatrix} = 0$$\alpha(1-0) - 2(1-0) + \beta(1-0) = 0 \Rightarrow \alpha - 2 + \beta = 0 \Rightarrow \alpha + \beta = 2 \dots (i)$Since $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$,$\vec{a}$ must be proportional to the sum of the unit vectors along $\vec{b}$ and $\vec{c}$:$\hat{b} = \frac{\vec{b}}{|\vec{b}|} = \frac{(1, 1, 0)}{\sqrt{2}}, \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{(0, 1, 1)}{\sqrt{2}}$$\vec{a} = k(\hat{b} + \hat{c}) = \frac{k}{\sqrt{2}}(1, 2, 1)$Comparing this with $\vec{a} = (\alpha, 2, \beta)$,we get $\frac{k}{\sqrt{2}} = 1$,so $k = \sqrt{2}$.Thus,$\alpha = 1$ and $\beta = 1$.

The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

Similar Questions

If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{r}$ is a non-zero vector such that $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$,then $\vec{r} = $

$\vec{a}, \vec{b}, \text{ and } \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. If $\vec{a}, \vec{b}, \vec{c}$ are perpendicular to the vectors $\vec{b}+\vec{c}, \vec{c}+\vec{a}, \vec{a}+\vec{b}$ respectively,then $\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2-2} = $

Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals:

Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.
Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$

Vedclass Test Series

Exam Paper Generator

Online Exam Module