Let the equation of the plane P containing th | vedclass

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

Question

(C) The line is given by $\frac{x+10}{1} = \frac{y-8}{-2} = \frac{z}{1}$. The point on the line is $(-10, 8, 0)$ and the direction ratios are $(1, -2,

Vedclass · Accepted Answer

$355$

Vedclass · Answer

(C) The line is given by $\frac{x+10}{1} = \frac{y-8}{-2} = \frac{z}{1}$. The point on the line is $(-10, 8, 0)$ and the direction ratios are $(1, -2, 1)$.Since the plane $ax + by + 3z = 2(a+b)$ contains the point $(-10, 8, 0)$,we have $a(-10) + b(8) + 3(0) = 2a + 2b$,which simplifies to $-10a + 8b = 2a + 2b$,so $6b = 12a$,or $b = 2a$.Since the normal to the plane $(a, b, 3)$ is perpendicular to the line with direction ratios $(1, -2, 1)$,their dot product is zero: $a(1) + b(-2) + 3(1) = 0$,so $a - 2b + 3 = 0$.Substituting $b = 2a$ into the equation $a - 2b + 3 = 0$,we get $a - 2(2a) + 3 = 0$,which gives $-3a = -3$,so $a = 1$ and $b = 2$.The equation of the plane is $x + 2y + 3z = 2(1+2) = 6$,or $x + 2y + 3z - 6 = 0$.The distance $c$ from the point $(1, 27, 7)$ to the plane is $c = \frac{|1(1) + 2(27) + 3(7) - 6|}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{|1 + 54 + 21 - 6|}{\sqrt{14}} = \frac{70}{\sqrt{14}} = 5\sqrt{14}$.Thus,$c^2 = 25 	imes 14 = 350$.Finally,$a^2 + b^2 + c^2 = 1^2 + 2^2 + 350 = 1 + 4 + 350 = 355$.

Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ and the distance of the plane $P$ from the point $(1,27,7)$ be $c$. Then $a^2+b^2+c^2$ is equal to $.............$.

Similar Questions

The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

If the line $\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}$ lies in the plane $3x-14y+6z+49=0$,then the value of $m$ is

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.

If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module