From a point P(lambda, lambda, lambda),perpen | vedclass

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

Question

(C) The first line is $L_1: \frac{x}{1} = \frac{y}{1}, z=1$. Let any point $Q$ on $L_1$ be $(\alpha, \alpha, 1)$.Direction ratios of $PQ$ are $(\alpha

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) The first line is $L_1: \frac{x}{1} = \frac{y}{1}, z=1$. Let any point $Q$ on $L_1$ be $(\alpha, \alpha, 1)$.Direction ratios of $PQ$ are $(\alpha-\lambda, \alpha-\lambda, 1-\lambda)$.Since $PQ$ is perpendicular to $L_1$ (direction vector $(1, 1, 0)$),we have $(\alpha-\lambda)(1) + (\alpha-\lambda)(1) + (1-\lambda)(0) = 0$,which gives $2(\alpha-\lambda) = 0$,so $\alpha = \lambda$.Thus,$Q = (\lambda, \lambda, 1)$ and the vector $\vec{PQ} = (0, 0, 1-\lambda)$.The second line is $L_2: \frac{x}{-1} = \frac{y}{1}, z=-1$. Let any point $R$ on $L_2$ be $(-\beta, \beta, -1)$.Direction ratios of $PR$ are $(-\beta-\lambda, \beta-\lambda, -1-\lambda)$.Since $PR$ is perpendicular to $L_2$ (direction vector $(-1, 1, 0)$),we have $(-\beta-\lambda)(-1) + (\beta-\lambda)(1) + (-1-\lambda)(0) = 0$,which gives $\beta+\lambda+\beta-\lambda = 0$,so $2\beta = 0$,hence $\beta = 0$.Thus,$R = (0, 0, -1)$ and the vector $\vec{PR} = (-\lambda, -\lambda, -1-\lambda)$.Given $\angle QPR = 90^\circ$,so $\vec{PQ} \cdot \vec{PR} = 0$.$(0)(-\lambda) + (0)(-\lambda) + (1-\lambda)(-1-\lambda) = 0$.$-(1-\lambda)(1+\lambda) = 0 \Rightarrow \lambda^2 - 1 = 0 \Rightarrow \lambda = \pm 1$.If $\lambda = 1$,$P = (1, 1, 1)$,which lies on $L_1$,so $PQ$ is not uniquely defined. Thus,$\lambda = -1$ is the only solution.

From a point $P(\lambda, \lambda, \lambda)$,perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x, z=1$ and $y=-x, z=-1$. If $P$ is such that $\angle QPR$ is a right angle,then the possible value$(s)$ of $\lambda$ is(are)

Similar Questions

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

The distance of the point $(3, 4, 5)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=2$ is: (in $units$)

If the angle $\theta$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2x - y + \sqrt{\lambda} z + 4 = 0$ is such that $\sin \theta = \frac{1}{3}$,the value of $\lambda$ is

Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

Let the plane $2x + 3y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3y + 5z = 8$. If the mirror image of the point $(2, -1/2, 2)$ in the rotated plane is $B(a, b, c)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module