Let the plane x+3y-2z+6=0 meet the coordinate | vedclass

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

Question

(D) The plane equation is $x+3y-2z+6=0$. Setting two coordinates to zero,we find the intercepts: $A(-6, 0, 0)$,$B(0, -2, 0)$,$C(0, 0, 3)$. Let $H(\alp

Vedclass · Accepted Answer

$288$

Vedclass · Answer

(D) The plane equation is $x+3y-2z+6=0$. Setting two coordinates to zero,we find the intercepts: $A(-6, 0, 0)$,$B(0, -2, 0)$,$C(0, 0, 3)$. Let $H(\alpha, \beta, \frac{6}{7})$ be the orthocentre. Since $H$ lies on the plane $ABC$,$\alpha + 3\beta - 2(\frac{6}{7}) + 6 = 0 \implies \alpha + 3\beta = -6 + \frac{12}{7} = -\frac{30}{7}$. Also,$\overrightarrow{AH} \cdot \overrightarrow{BC} = 0$. $\overrightarrow{AH} = (\alpha+6, \beta, \frac{6}{7}-0) = (\alpha+6, \beta, \frac{6}{7})$. $\overrightarrow{BC} = (0, 2, 3)$. $(\alpha+6)(0) + \beta(2) + \frac{6}{7}(3) = 0 \implies 2\beta + \frac{18}{7} = 0 \implies \beta = -\frac{9}{7}$. Substituting $\beta$ into the plane equation: $\alpha + 3(-\frac{9}{7}) = -\frac{30}{7} \implies \alpha - \frac{27}{7} = -\frac{30}{7} \implies \alpha = -\frac{3}{7}$. Now,$98(\alpha+\beta)^2 = 98(-\frac{3}{7} - \frac{9}{7})^2 = 98(-\frac{12}{7})^2 = 98 	imes \frac{144}{49} = 2 	imes 144 = 288$.

Let the plane $x+3y-2z+6=0$ meet the coordinate axes at the points $A, B, C$. If the orthocentre of the triangle $ABC$ is $\left(\alpha, \beta, \frac{6}{7}\right)$,then $98(\alpha+\beta)^2$ is equal to $........$.

Similar Questions

The vector equation of the plane which is at a distance of $ \frac{3}{\sqrt{14}} $ from the origin and the normal vector from the origin is $ 2 \hat{i}-3 \hat{j}+\hat{k} $ is:

The equation of the plane passing through the points $(-1, -2, 0)$ and $(2, 3, 5)$ and parallel to the line $r = -3j + k + \lambda(2i + 5j - k)$ is

$A$ plane meets the coordinate axes at the points $A, B, C$ respectively in such a way that the centroid of $\triangle ABC$ is $(1, r, r^2)$ for some real $r$. If the plane passes through the point $(5, 5, -12)$, then $r=$

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.

Find the equation of the plane bisecting the angle between the planes $2x - y + 2z + 3 = 0$ and $3x - 2y + 6z + 8 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module