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VITEEE 2025 Paper

Question : 104 of 125

Marks: +1, -0

(1,2,-4)

\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}

\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}

Solution:

Let

a, b, c

, be the direction ratios of the required line. Then, equation of line passing through

(1,2,-4)

and having DR's (

a, b, c

) is

\frac{x-1}{a} = \frac{y-2}{b} = \frac{z+4}{c}

Now, as line (i) is perpendicular to the lines

\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}

and

\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}

\therefore 3a-16b+7c=0

3a+8b-5c=0

On solving (ii) and (iii), we get

\frac{a}{24} = \frac{b}{36} = \frac{c}{72} \; \text{i.e.,}\; \frac{a}{2} = \frac{b}{3} = \frac{c}{6}

So, the required equation is

\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}

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