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UPSC NDA I 2014 Mathematics Paper

For the next three (03) items that follow:

A straight line passes through (1, -2, 3) and perpendicular to the plane

2x + 3y – z = 7

[NDA I 2014]

Question : 62 of 120

Marks: +1, -0

Solution:

Since we have the plane

2x+3y-z=7,

we know that the vector is

[2,3,-1]

perpendicular to the plane

With the point

(1,-2,3)

we get the parametric equations for the line

x=1+2t,y=-2+3t,z=3-t

The line equation can be found by solving for t

t=\frac{x-1}{2}=\frac{y+2}{3}=-\frac{z-3}{1}

For intersection point

2(1+2t)+3(-2+3t)-(3-t)=7

\Rightarrow 2+4t-6+9t-3+t=7

\Rightarrow t=1

Therefore, the point is

(1+2,-2+3,3-1)\equiv(3,1,2)

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