If the vector bar{i}-7 bar{j}+2 bar{k} is alo | vedclass

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

Question

(B) Let $\bar{u} = \bar{i} - 7\bar{j} + 2\bar{k}$ be the vector along the internal bisector. The magnitude $|\bar{u}| = \sqrt{1^2 + (-7)^2 + 2^2} = \s

Vedclass · Accepted Answer

$\frac{7}{9}$

Vedclass · Answer

(B) Let $\bar{u} = \bar{i} - 7\bar{j} + 2\bar{k}$ be the vector along the internal bisector. The magnitude $|\bar{u}| = \sqrt{1^2 + (-7)^2 + 2^2} = \sqrt{1 + 49 + 4} = \sqrt{54} = 3\sqrt{6}$.Let $\bar{b} = -2\bar{i} - \bar{j} + 2\bar{k}$. The magnitude $|\bar{b}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = 3$.The unit vector along $\bar{b}$ is $\hat{b} = \frac{-2\bar{i} - \bar{j} + 2\bar{k}}{3} = -\frac{2}{3}\bar{i} - \frac{1}{3}\bar{j} + \frac{2}{3}\bar{k}$.Let $\hat{a} = x\bar{i} + y\bar{j} + z\bar{k}$ be the unit vector along $\bar{a}$.The internal bisector is along $\hat{a} + \hat{b}$. Thus,$\hat{a} + \hat{b} = k\bar{u}$ for some scalar $k > 0$.$\hat{a} = k(\bar{i} - 7\bar{j} + 2\bar{k}) - (-\frac{2}{3}\bar{i} - \frac{1}{3}\bar{j} + \frac{2}{3}\bar{k}) = (k + \frac{2}{3})\bar{i} + (-7k + \frac{1}{3})\bar{j} + (2k - \frac{2}{3})\bar{k}$.Since $|\hat{a}| = 1$,we have $(k + \frac{2}{3})^2 + (-7k + \frac{1}{3})^2 + (2k - \frac{2}{3})^2 = 1$.$(k^2 + \frac{4}{3}k + \frac{4}{9}) + (49k^2 - \frac{14}{3}k + \frac{1}{9}) + (4k^2 - \frac{8}{3}k + \frac{4}{9}) = 1$.$54k^2 - 6k + 1 = 1 \implies 54k^2 - 6k = 0 \implies 6k(9k - 1) = 0$.Since $k > 0$,$k = \frac{1}{9}$.Then $x = k + \frac{2}{3} = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}$.

If the vector $\bar{i}-7 \bar{j}+2 \bar{k}$ is along the internal bisector of the angle between the vectors $\bar{a}$ and $-2 \bar{i}-\bar{j}+2 \bar{k}$ and the unit vector along $\bar{a}$ is $x \bar{i}+y \bar{j}+z \bar{k}$ then $x=$

Similar Questions

If $A \equiv (2i + 3j)$, $B \equiv (\pi + 9j)$, and $C \equiv (i - j)$ are collinear, then the value of $p$ is: (in $/2$)

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $O$ is any point,then $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = $

If $ABCDEF$ is a regular hexagon,where two adjacent sides $\vec{AB}$ and $\vec{BC}$ are $\vec{a}$ and $\vec{b}$ respectively. Then $\vec{CD}$ is

Let $OA = a, OB = b$ be two non-collinear vectors, $OP = x_1 a + y_1 b, OQ = x_2 a + y_2 b$ and $A^{\prime}O = OA, B^{\prime}O = OB$. If $x_1 = -\frac{3}{4}, x_2 = \frac{1}{3}, y_1 = \frac{7}{4}, y_2 = \frac{5}{3}$, then

The non-zero vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are related by $\vec{a} = 8\vec{b}$ and $\vec{c} = -7\vec{b}$. Then the angle between $\vec{a}$ and $\vec{c}$ is ............... $^\circ $

Vedclass Test Series

Exam Paper Generator

Online Exam Module