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

$A$ plane $\Pi$ passes through the points $A=(0,0,2)$, $B=(1,0,1)$, and $C=(3,1,1)$. If the plane $\Pi$ makes angles $\alpha$ and $\beta$ with the $XY$ and $XZ$-coordinate planes respectively, then $\sin^2 \alpha + \sin^2 \beta =$

Let the mirror image of the point $P(1, 3, a)$ with respect to the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - b = 0$ be $Q(-3, 5, 2)$. Then the value of $|a + b|$ is equal to ......

The equation of the plane passing through the points $(1, 2, 3)$,$(-1, 4, 2)$ and $(3, 1, 1)$ is

$A$ plane meets the coordinate axes at points $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, r, r^2)$. Find the equation of the plane.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module