UPSC NDA Math Model Paper 2

Given: Equation of line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{1}$ and equation of plane is x - y + z - 5 = 0 First let's find out if they given line is parallel to the plane or not. As we can see that, the direction ratios of the given line are: (1, 2, 1) Similarly, the direction ratios of the normal to the given plane are: (1, -1, 1) ⇒ 1 × 1 - 2 × 1 + 1 × 1 = 0 So, the given line is parallel to the given plane. Let's find out a point on the given line As we can see that, point Q (3, 4, 5) lies on the line. So, finding the distance between the point Q and the given plane is same as finding the distance between the given line and plane As we know that, distance between a point and a plane is given by: $\left|\frac{A x_{1}+B y_{1}+C z_{1}-d}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|$ Here, $x_{1}=3, y_{1}=4$ and $z_{1}=5$ $\Rightarrow d=\left|\frac{3 \times 1+4 \times (-1)+5 \times 1-5}{\sqrt{1^{2}+(-1)^{2}+1^{2}}}\right|$ $=\frac{1}{\sqrt{3}}$ units Hence, option B is the correct answer.