The line drawn from (4, -1, 2) to the point ( | vedclass

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

Question

(A) The line passes through the points $P(4, -1, 2)$ and $Q(-3, 2, 3)$.The direction ratios of this line are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_

Vedclass · Accepted Answer

$7x - 3y - z + 89 = 0$

Vedclass · Answer

(A) The line passes through the points $P(4, -1, 2)$ and $Q(-3, 2, 3)$.The direction ratios of this line are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (-3 - 4, 2 - (-1), 3 - 2) = (-7, 3, -1)$.Since the line is perpendicular to the plane,the direction ratios of the normal to the plane are the same as the direction ratios of the line,which are $(-7, 3, -1)$.Alternatively,we can use the normal vector $\vec{n} = (7, -3, 1)$ by multiplying by $-1$.The plane passes through the point $R(-10, 5, 4)$.The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.Substituting the values,we get $7(x - (-10)) - 3(y - 5) - 1(z - 4) = 0$.$7(x + 10) - 3(y - 5) - (z - 4) = 0$.$7x + 70 - 3y + 15 - z + 4 = 0$.$7x - 3y - z + 89 = 0$.

The line drawn from $(4, -1, 2)$ to the point $(-3, 2, 3)$ meets a plane at right angles at the point $(-10, 5, 4)$. Find the equation of the plane.

Similar Questions

The direction cosines of a normal to the plane passing through $(4,2,3)$,$(-1,4,2)$ and $(3,2,1)$ are .....

The foot of the perpendicular drawn from the origin to a plane is $M(2, 1, -2)$. Find the vector equation of the plane.

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the axes at points $A, B,$ and $C$. Find the area of $\Delta ABC$.

Find the equation of the set of points which are equidistant from the points $A(3, 4, -5)$ and $B(-2, 1, 4)$.

If the foot of the perpendicular drawn from the origin to a plane is $(1, 2, 3)$,then a point on that plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module