Let the plane ax + by + cz = d pass through ( | vedclass

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

Question

(D) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\vec{n_1} = 2\hat{i} + \hat{j} - 5\h

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(D) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors of the given planes,$\vec{n_1} = 2\hat{i} + \hat{j} - 5\hat{k}$ and $\vec{n_2} = 3\hat{i} + 5\hat{j} - 7\hat{k}$.Thus,$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -5 \ 3 & 5 & -7 \end{vmatrix} = \hat{i}(-7 + 25) - \hat{j}(-14 + 15) + \hat{k}(10 - 3) = 18\hat{i} - \hat{j} + 7\hat{k}$.The equation of the plane is $18x - y + 7z = d$.Since the plane passes through $(2, 3, -5)$,we have $18(2) - (3) + 7(-5) = d$,which gives $36 - 3 - 35 = d$,so $d = -2$.The equation of the plane is $18x - y + 7z = -2$,or $-18x + y - 7z = 2$.Comparing this with $ax + by + cz = d$,we get $a = -18, b = 1, c = -7, d = 2$.Since $	ext{gcd}(|-18|, |1|, |-7|, 2) = 1$ and $d > 0$,these values are correct.Finally,$a + 7b + c + 20d = -18 + 7(1) + (-7) + 20(2) = -18 + 7 - 7 + 40 = 22$.

Let the plane $ax + by + cz = d$ pass through $(2, 3, -5)$ and be perpendicular to the planes $2x + y - 5z = 10$ and $3x + 5y - 7z = 12$. If $a, b, c, d$ are integers,$d > 0$,and $\text{gcd}(|a|, |b|, |c|, d) = 1$,then the value of $a + 7b + c + 20d$ is equal to

Similar Questions

$A$ plane passes through $(2,3,-1)$ and is perpendicular to the line having direction ratios $3,-4,7$. The perpendicular distance from the origin to this plane is

$A$ plane passes through a fixed point $A(1, -2, 3)$. If the locus of the foot of the perpendicular to it from the point $B(0, 1, 2)$ is $x^2 + y^2 + z^2 + \alpha x + \beta y + \gamma z + \delta = 0$,then the value of $\alpha + \beta + \gamma + \delta$ is

Find the equation of the plane that passes through the three points $(1, 1, -1)$,$(6, 4, -5)$,and $(-4, -2, 3)$.

$A$ plane $\pi$ passing through the point $3 \hat{i}-4 \hat{j}+5 \hat{k}$ is parallel to the plane which passes through the point $\hat{i}+\hat{j}-\hat{k}$ and is perpendicular to the vector $\hat{i}+2 \hat{j}-3 \hat{k}$. Then the Cartesian equation of $\pi$ is

$P$ and $Q$ are points on the straight line passing through the point $A(3 \hat{i}+\hat{j}-\hat{k})$ and parallel to the vector $\vec{v} = 2 \hat{i}-\hat{j}+2 \hat{k}$. If $AP = AQ = 3$,then the vector equation of the plane $OPQ$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module