Let P be the plane passing through the point | vedclass

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

Question

(C) The direction vector of the line joining $(4, 1, -3)$ and $(2, 4, 3)$ is given by $\vec{n} = (2-4, 4-1, 3-(-3)) = (-2, 3, 6)$.Since the plane $P$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The direction vector of the line joining $(4, 1, -3)$ and $(2, 4, 3)$ is given by $\vec{n} = (2-4, 4-1, 3-(-3)) = (-2, 3, 6)$.Since the plane $P$ is perpendicular to this line,the normal vector to the plane is $\vec{n} = (-2, 3, 6)$.The equation of the plane passing through $(1, -1, -5)$ with normal vector $\vec{n} = (-2, 3, 6)$ is:$-2(x - 1) + 3(y + 1) + 6(z + 5) = 0$$-2x + 2 + 3y + 3 + 6z + 30 = 0$$-2x + 3y + 6z + 35 = 0$ or $2x - 3y - 6z = 35$.The distance of the point $(3, -2, 2)$ from the plane $2x - 3y - 6z - 35 = 0$ is given by the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.$d = \frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{2^2 + (-3)^2 + (-6)^2}}$$d = \frac{|6 + 6 - 12 - 35|}{\sqrt{4 + 9 + 36}}$$d = \frac{|-35|}{\sqrt{49}} = \frac{35}{7} = 5$.

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is

Similar Questions

If the equation of the plane bisecting the line segment joining the points $P(3,2,4)$ and $Q(-1,0,-2)$ and perpendicular to $PQ$ is $ax+by+cz+d=0$,then find the value of $ac+bd$.

$A$ plane which is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$ passes through $(1, 2, 1)$. The distance of the plane from the point $(2, 3, 4)$ is

The equation $axy + byz = cy$ represents the locus of the points which lie on

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1), (1, -\lambda^2, 1)$ and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to

Let $\alpha x+\beta y+\gamma z=1$ be the equation of a plane passing through the point $(3, -2, 5)$ and perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 5)$. Then the value of $\alpha \beta \gamma$ is equal to $..........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module