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  • Let P1 be the plane 3x -y -7z = 11 and P2 be the plane passing through the points (2, -1,0), (2,0, -1), and (5, 1, 1). If the foot of the perpendicular drawn from the point (7, 4, 1) on the

Let P1 be the plane 3x -y -7z = 11 and P2 be the plane passing through the points (2, -1,0), (2,0, -1), and (5, 1,
1). If the foot of the perpendicular drawn from the point (7, 4, 1) on the line of intersection of the planes P1
and P2 is (\alpha ,\beta ,\gamma), then \alpha + \beta +\gammais equal to _____.

Option: 1

11


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

P_2 is given by \left|\begin{array}{ccc} x-5 & y-1 & z-1 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right|=0 x-y-z=3 DR of line intersection of \mathrm{P}_1 \& \mathrm{P}_2 \begin{aligned} & \left|\begin{array}{ccc} i & j & k \\ 1 & -1 & 1 \\ 3 & -1 & -7 \end{array}\right| \\ & +6 \hat{i}+4 \hat{j}+2 \hat{k} \end{aligned} Let \begin{gathered} z=0, \quad x-y=3 \\ 3 x-y=11 \\ 2 x=8 \\ x=4 \\ y=1 \end{gathered}

 

So Line is $$ \begin{aligned} & \qquad \frac{\mathrm{x}-4}{6}=\frac{\mathrm{y}-1}{4}=\frac{\mathrm{z}-0}{2}=\mathrm{r} \\ & (\alpha, \beta, \gamma)=(6 \mathrm{r}+4,4 \mathrm{r}+1,2 \mathrm{r}) \\ & 6(\alpha-7)+4(\beta-4)+2(\gamma+1)=0 \\ & 6 \alpha-42+4 \beta-16+2 \gamma+2=0 \\ & 36 \mathrm{r}+24+16 \mathrm{r}+4+4 \mathrm{r}-56=0 \\ & 56 \mathrm{r}=28 \\ & \mathrm{r}=\frac{1}{2} \\ & \begin{array}{c} \alpha+\beta+\gamma=12 \mathrm{r}+5 \\ \quad=6+5=11 \end{array} \end{aligned}

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