A point P(a, a, a) lies on a line passing thr | vedclass

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

Question

(A) The line passes through the origin and makes equal angles with the axes,so its direction ratios are $(1, 1, 1)$.The equation of the line is $\frac

Vedclass · Accepted Answer

$1/a$

Vedclass · Answer

(A) The line passes through the origin and makes equal angles with the axes,so its direction ratios are $(1, 1, 1)$.The equation of the line is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$.The point $P$ is $(a, a, a)$.The plane passes through $P(a, a, a)$ and is perpendicular to $OP$. The direction ratios of the normal to the plane are the same as the direction ratios of the line $OP$,which are $(1, 1, 1)$.The equation of the plane is $1(x - a) + 1(y - a) + 1(z - a) = 0$.Simplifying this,we get $x + y + z = 3a$.Dividing by $3a$,we get $\frac{x}{3a} + \frac{y}{3a} + \frac{z}{3a} = 1$.The intercepts on the axes are $3a, 3a, 3a$.The sum of the reciprocals of the intercepts is $\frac{1}{3a} + \frac{1}{3a} + \frac{1}{3a} = \frac{3}{3a} = \frac{1}{a}$.

$A$ point $P(a, a, a)$ lies on a line passing through the origin and making equal angles with the axes,where $a$ is a constant. $A$ plane passes through $P$ and is perpendicular to $OP$. If this plane makes intercepts on the axes,what is the sum of the reciprocals of these intercepts?

Similar Questions

If the foot of the perpendicular drawn from the point $(1,0,-2)$ to the plane $\pi$ is $(2,0,-1)$ and the equation of the plane $\pi$ is $ax+by+cz=2$, then $a^2+b^2+c^2=$

$A$ plane containing two lines whose direction ratios are $(-1, 2, 1)$ and $(1, 3, 2)$ passes through the point $(2, 1, k)$. If this plane also passes through the point $(3, -1, 4)$, then $k=$

The equation of the plane,passing through the point $(1, 1, 1)$ and perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,is

If a plane $\pi$ passes through the point $(-1,6,2)$ and is perpendicular to the planes $x+2y+2z-5=0$ and $3x+3y+2z-8=0$, then the perpendicular distance from the point $(1,-1,1)$ to the plane $\pi$ is

$A$ plane $\Pi$ passes through the points $A=(0,0,2)$, $B=(1,0,1)$, and $C=(3,1,1)$. If the plane $\Pi$ makes angles $\alpha$ and $\beta$ with the $XY$ and $XZ$-coordinate planes respectively, then $\sin^2 \alpha + \sin^2 \beta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module