The foot of the perpendicular drawn from a po | vedclass

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

Question

(A) The foot of the perpendicular from $A(1,1,1)$ to the plane $\pi$ is $P(-3,3,5)$. The vector $\vec{AP} = P - A = (-3-1, 3-1, 5-1) = (-4, 2, 4)$ is

Vedclass · Accepted Answer

$-10$

Vedclass · Answer

(A) The foot of the perpendicular from $A(1,1,1)$ to the plane $\pi$ is $P(-3,3,5)$. The vector $\vec{AP} = P - A = (-3-1, 3-1, 5-1) = (-4, 2, 4)$ is the normal to the plane $\pi$. Since the plane $\pi$ passes through $P(-3,3,5)$,its equation is $-4(x+3) + 2(y-3) + 4(z-5) = 0$,which simplifies to $-4x + 2y + 4z - 38 = 0$,or $2x - y - 2z + 19 = 0$. The midpoint $M$ of $AP$ is $(\frac{1-3}{2}, \frac{1+3}{2}, \frac{1+5}{2}) = (-1, 2, 3)$. $A$ plane parallel to $\pi$ has the form $2x - y - 2z + k = 0$. Since it passes through $M(-1, 2, 3)$,we have $2(-1) - (2) - 2(3) + k = 0$,which gives $-2 - 2 - 6 + k = 0$,so $k = 10$. The equation is $2x - y - 2z + 10 = 0$. Comparing this with $ax - y + cz + d = 0$,we get $a = 2$,$c = -2$,and $d = 10$. Thus,$a + c - d = 2 + (-2) - 10 = -10$.

The foot of the perpendicular drawn from a point $A(1,1,1)$ onto a plane $\pi$ is $P(-3,3,5)$. If the equation of the plane parallel to the plane $\pi$ and passing through the midpoint of $AP$ is $ax-y+cz+d=0$,then $a+c-d=$

Similar Questions

If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the plane $2x + 3y - z + 13 = 0$ at a point $P$ and the plane $3x + y + 4z = 16$ at a point $Q$,then $PQ$ is equal to

The distance of the point $(1, 6, 2)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=16$ is (in $\text{ units}$)

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to each of the planes $3x - 2y + z = 8$ and $x + y + z = 6$ is $lx + my + nz = 1$,then $4m + 2n - 31 =$

The equation of the plane containing the line $2x - 5y + z = 3; x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is:

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 .... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module