If for a 0, the feet of perpendiculars from | vedclass

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

Question

(D) Since $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0,$ we have $l(0) + m(-a) + n(-1) = 0,$ which implies $-ma - n = 0,$ or $\frac{m}{n} = -\fr

Vedclass · Accepted Answer

$\sqrt{66}$

Vedclass · Answer

(D) Since $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0,$ we have $l(0) + m(-a) + n(-1) = 0,$ which implies $-ma - n = 0,$ or $\frac{m}{n} = -\frac{1}{a} \quad \dots(1)$The vector $\vec{CA} = (a - 0, -2a - (-a), 3 - (-1)) = (a, -a, 4)$ is parallel to the normal vector of the plane $\vec{n} = (l, m, n).$Thus,$\frac{a}{l} = \frac{-a}{m} = \frac{4}{n}.$ From $\frac{-a}{m} = \frac{4}{n},$ we get $\frac{m}{n} = -\frac{a}{4} \quad \dots(2)$Equating $(1)$ and $(2),$ we have $-\frac{1}{a} = -\frac{a}{4} \Rightarrow a^2 = 4.$ Since $a > 0,$ we have $a = 2.$Substituting $a = 2$ into $(1),$ we get $\frac{m}{n} = -\frac{1}{2}.$ Let $m = -t$ and $n = 2t.$ Then $\frac{2}{l} = \frac{-2}{-t} \Rightarrow l = t.$The equation of the plane is $t(x - y + 2z) = 0,$ or $x - y + 2z = 0.$The point $D$ is the foot of the perpendicular from $B(0, 4, 5)$ to the plane $x - y + 2z = 0.$ The line $BD$ is given by $\frac{x-0}{1} = \frac{y-4}{-1} = \frac{z-5}{2} = k.$So,$D = (k, 4-k, 5+2k).$ Since $D$ lies on the plane,$k - (4-k) + 2(5+2k) = 0 \Rightarrow k - 4 + k + 10 + 4k = 0 \Rightarrow 6k + 6 = 0 \Rightarrow k = -1.$Thus,$D = (-1, 4-(-1), 5+2(-1)) = (-1, 5, 3).$With $C = (0, -2, -1)$ and $D = (-1, 5, 3),$ the length $CD = \sqrt{(-1-0)^2 + (5-(-2))^2 + (3-(-1))^2} = \sqrt{(-1)^2 + 7^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}.$

If for $a > 0,$ the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively,then the length of line segment $CD$ is equal to

Similar Questions

The acute angle between the planes $P_{1}$ and $P_{2}$,when $P_{1}$ and $P_{2}$ are the planes passing through the intersection of the planes $5x + 8y + 13z - 29 = 0$ and $8x - 7y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$,respectively,is

The equation of the line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is:

The length (in units) of the projection of the line segment,joining the points $(5,-1,4)$ and $(4,-1,3)$,on the plane $x+y+z=7$ is

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is $.........$.

$A$ point $P$ lies on a line passing through $Q(1, -2, 3)$ and is parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$. If $P$ lies on the plane $2x + 3y - 4z + 22 = 0$, then the length of the segment $PQ$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module