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VITEEE 2017 Paper

Section: Aptitude

Question : 112 of 125

Marks: +1, -0

\frac{x}{2}

\frac{y}{3}

\frac{z-1}{-6}

Solution:

Equation of the line through (1, -2, 3) parallel to the line

\frac{x}{2}

\frac{y}{3}

\frac{z-1}{-6}

\frac{x-1}{2}

\frac{y+2}{3}

\frac{z-1}{-6}

= r (say) ... (1)

Then any point on (1) is (2r + 1,3r -2, -6r + 3)

If this point lies on the plane x - y + z = 5 then

(2r + 1) - (3r - 2) + (-6r + 3) = 2 ⇒ r =

\frac{1}{7}

Hence the point is

\left(97, \frac{-11}{7}, \frac{15}{7}\right)

Distance between (1, -2, 3) and

\left(97, \frac{-11}{7}, \frac{15}{7}\right)

\sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}}

\sqrt{\frac{49}{49}}

= 1

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