If (3,4,-7) is the foot of the perpendicular | vedclass

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(A) The normal vector to the plane $\pi$ is the vector joining the point $P(-2,3,6)$ and the foot of the perpendicular $F(3,4,-7)$.Thus,the direction

Vedclass · Accepted Answer

$132$

Vedclass · Answer

(A) The normal vector to the plane $\pi$ is the vector joining the point $P(-2,3,6)$ and the foot of the perpendicular $F(3,4,-7)$.Thus,the direction ratios of the normal are $(3 - (-2), 4 - 3, -7 - 6) = (5, 1, -13)$.The equation of the plane passing through $(3,4,-7)$ with normal vector $(5, 1, -13)$ is given by $5(x - 3) + 1(y - 4) - 13(z + 7) = 0$.$5x - 15 + y - 4 - 13z - 91 = 0$$5x + y - 13z = 110$.To find the intercepts,we write the equation in intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$:$\frac{5x}{110} + \frac{y}{110} - \frac{13z}{110} = 1$$\frac{x}{22} + \frac{y}{110} + \frac{z}{-\frac{110}{13}} = 1$.The $X$-intercept is $a = 22$ and the $Y$-intercept is $b = 110$.The sum of the $X$ and $Y$-intercepts is $22 + 110 = 132$.

If $(3,4,-7)$ is the foot of the perpendicular drawn from the point $(-2,3,6)$ to the plane $\pi$,then the sum of the intercepts made by the plane $\pi$ on the $X$ and $Y$-axes is

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