The direction ratios of a bisector of the ang | vedclass

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

Question

(A) Let the vectors along the two lines be $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \sqrt{3}\hat{i} - \sqrt{3}\hat{j} + 0\hat{k}$.First

Vedclass · Accepted Answer

$1+\sqrt{3}, 1-\sqrt{3}, 2$

Vedclass · Answer

(A) Let the vectors along the two lines be $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \sqrt{3}\hat{i} - \sqrt{3}\hat{j} + 0\hat{k}$.First, calculate the magnitudes of the vectors:$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.$|\vec{b}| = \sqrt{(\sqrt{3})^2 + (-\sqrt{3})^2 + 0^2} = \sqrt{3 + 3 + 0} = \sqrt{6}$.Since the magnitudes are equal $(|\vec{a}| = |\vec{b}|)$, the direction ratios of the angle bisector are given by the components of the vector $\vec{a} + \vec{b}$ or $\vec{a} - \vec{b}$.Calculating $\vec{a} + \vec{b} = (1 + \sqrt{3})\hat{i} + (1 - \sqrt{3})\hat{j} + (2 + 0)\hat{k}$.Thus, the direction ratios are $(1 + \sqrt{3}, 1 - \sqrt{3}, 2)$.Comparing this with the given options, option $A$ is correct.

The direction ratios of a bisector of the angle between two lines whose direction ratios are $1, 1, 2$ and $\sqrt{3}, -\sqrt{3}, 0$ are

Similar Questions

The point of intersection of the two lines $\frac{x - 3}{3} = \frac{y - 3}{-1}, z - 1 = 0$ and $\frac{x - 6}{2} = \frac{z - 1}{3}, y - 2 = 0$ is

The equation of the $x$-axis is:

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

The distance of the point $P(-3, 2, 3)$ from the line passing through $A(4, 6, -2)$ with direction ratios $\langle -1, 2, 3 \rangle$ is . . . . . . units.

The $xy$-plane divides the line segment joining the points $(2, 4, 5)$ and $(-4, 3, -2)$ in the ratio:

Vedclass Test Series

Exam Paper Generator

Online Exam Module