"If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is ($\alpha,\beta,\gamma$), then $\alpha+\beta+\gamma$ is equal to :"

Question

Accepted Answer

"Direction of line

$$ \begin{aligned} \vec{b} & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 0 & 3 & -1 \end{array}\right| \\\\ & =\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3) \\\\ & =-5 \hat{i}+\hat{j}+3 \hat{k} \end{aligned} $$

Equation of line

$\frac{x-3}{-5}=\frac{y-2}{1}=\frac{z-1}{3}$

Let foot of perpendicular be $=(-5 k+3, k+2,3 k+1)$

$\Rightarrow(-5 k+2)(-5)+(k-7)(1)+(3 k-6)(3)=0$

Or $25 k-10+k-7+9 k-18=0$

Or $k=1$

$\alpha+\beta+\gamma=-k+6=5$"

If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is ($\alpha,\beta,\gamma$), then $\alpha+\beta+\gamma$ is equal to :

Solution

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If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is ($\alpha,\beta,\gamma$), then $\alpha+\beta+\gamma$ is equal to :

Solution

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