Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The direction ratios of the line \(\frac{1-x}{3}=\frac{7y-14}{2}=\frac{z-3}{2}\)are

Options:

-3,\(\frac{1}{7}\),2

-3,2,2

-3,\(\frac{2}{7}\),1

-21,2,14

Correct Answer:

-21,2,14

Explanation:

\(=\frac{1-x}{3}=\frac{7y-14}{2}=\frac{z-3}{2}\)

\(\frac{-(x+1}{3}=\frac{(y-2)}{2}=\frac{z-3}{2}⇒\frac{z-1}{-3}=\frac{y-2}{\frac{2}{7}}=\frac{z-3}{2}\)

Direction ratio are 

x = -3

y = $\frac{2}{7}$

z = 2

So direction ratios are = (-3, $\frac{2}{7}$, 2) or $\frac{-3×7,2,2×7}{7}$

$=\frac{-21,2,14}{7}=(-21,2,14)$ taking same L.C.M

So option 4 is correct.