If the length of the perpendicular from the p | vedclass

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

Question

(B) Let the line be $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1} = \lambda$. Any point $C$ on the line is given by $(\lambda, 1, -\lambda - 1)$.

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let the line be $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1} = \lambda$. Any point $C$ on the line is given by $(\lambda, 1, -\lambda - 1)$. The given point is $P(\beta, 0, \beta)$. The direction vector of the line is $\vec{v} = (1, 0, -1)$. The vector $\vec{PC} = (\lambda - \beta, 1 - 0, -\lambda - 1 - \beta) = (\lambda - \beta, 1, -\lambda - \beta - 1)$. Since $PC$ is perpendicular to the line,$\vec{PC} \cdot \vec{v} = 0$: $(\lambda - \beta)(1) + (1)(0) + (-\lambda - \beta - 1)(-1) = 0$ $\lambda - \beta + \lambda + \beta + 1 = 0$ $2\lambda + 1 = 0 \implies \lambda = -\frac{1}{2}$. Thus,the point $C$ is $(-\frac{1}{2}, 1, -\frac{1}{2})$. The length of the perpendicular $PC$ is $\sqrt{(\beta - (-\frac{1}{2}))^2 + (0 - 1)^2 + (\beta - (-\frac{1}{2}))^2} = \sqrt{\frac{3}{2}}$. Squaring both sides: $(\beta + \frac{1}{2})^2 + 1 + (\beta + \frac{1}{2})^2 = \frac{3}{2}$. $2(\beta + \frac{1}{2})^2 = \frac{3}{2} - 1 = \frac{1}{2}$. $(\beta + \frac{1}{2})^2 = \frac{1}{4}$. $\beta + \frac{1}{2} = \pm \frac{1}{2}$. Case $1$: $\beta + \frac{1}{2} = \frac{1}{2} \implies \beta = 0$ (Rejected as $\beta 
eq 0$). Case $2$: $\beta + \frac{1}{2} = -\frac{1}{2} \implies \beta = -1$. Therefore,$\beta = -1$.

If the length of the perpendicular from the point $P(\beta, 0, \beta) \, (\beta \neq 0)$ to the line $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1}$ is $\sqrt{\frac{3}{2}}$,then $\beta$ is equal to

Similar Questions

The equation of the line passing through $(3, -1, 2)$ and perpendicular to the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\vec{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + \mu(\hat{i} - 2 \hat{j} + 2 \hat{k})$ is

Find the angle between the lines whose direction cosines are given by the equations $3l+m+5n=0$ and $6mn-2nl+5lm=0$.

The vector equation of the line $2x+4=3y+1=6z-3$ is

If the lines $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z + 1}{3}$ and $\frac{x + 2}{2} = \frac{y - k}{3} = \frac{z}{4}$ are coplanar,then the value of $k$ is

If the line joining the points $(k, 3, 4)$ and $(4, 7, 8)$ is parallel to the line joining the points $(-1, -2, 1)$ and $(1, 2, l)$,then $k + l =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module