Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

The equation of line passing through origin and parallel to the line $\vec r=3\hat i+4\hat j-5\hat k+t\left(2\hat i-\hat j+7\hat k\right)$, where t is a parameter, is:

(A) $\frac{x}{2}=\frac{y}{-1}=\frac{z}{7}$
(B) $\vec r=m(12\hat i- 6\hat j + 42\hat k)$; where m is the parameter
(C) $\vec r=(12\hat i-6\hat j+42\hat k)+s (0\hat i-0\hat j + 0\hat k)$; where s is the parameter
(D) $\frac{x-3}{0}=\frac{y-4}{0}=\frac{z+5}{0}$
(E) $\frac{x}{3}=\frac{y}{4}=\frac{z}{5}$

Choose the correct answer from the options given below:

Options:

(A) and (B) only

(A), (B) and (C) only

(C), (D) and (E) only

(A) only

Correct Answer:

(A) and (B) only

Explanation:

The correct answer is Option (1) → (A) and (B) only

The given equation of a line is,

$r=(3\hat i+4\hat j-5\hat k)+t(2\hat i-\hat j+7\hat k)$

∴ Direction vector, $d=2\hat i-\hat j+7\hat k$

Equation of line passing through origin,

$r=m(2\hat i-\hat j+7\hat k)$

$⇒\vec r=m(2\hat i-\hat j+7\hat k)$ (B)

the parametric form of the equation:

$x=2m,y=-m,z=7m$

$\frac{x}{2}=\frac{y}{-1}=\frac{z}{7}$ (A)