A. Equation of the line passing through the point (1, 2, 3) and parallel to the vector \(3\hat{\imath}+2\hat{\jmath}-2\hat{k}\) is \(\displaystyle\frac{x-1}{3}=\frac{y-2}{2}=\frac{y-3}{-2}\).
B. Equation of line passing through (1, 2, 3) and parallel to the line given by \(\displaystyle\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}\) is \(\displaystyle\frac{x-1}{3}=\frac{y-2}{5}=\frac{z+3}{6}\).
C. Equation of line passing through the origin and (5, −2, 3) is \(\displaystyle\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\).
D. Equation of plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratios 2, 3, −1 is \(2(x-1)+3(y-2)-1(z-3)=0\).
E. Equation of plane with intercepts 2, 3 and 4 on x, y and z-axis respectively is \(2x+3y+4z=1\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, C, D only
The line through \((x_1, y_1, z_1)\) with direction ratios a, b, c is
\( \displaystyle \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \)
The plane through \((x_1, y_1, z_1)\) with normal direction a, b, c is
\(a(x-x_1) + b(y-y_1) + c(z-z_1) = 0\)
The plane with intercepts p, q, r is
\( \displaystyle \frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1 \)
The line in (B) must first be written in standard form, with +y in the numerator:
\( \displaystyle \frac{x+3}{3} = \frac{y-4}{-5} = \frac{z+8}{6} \)
so its direction ratios are 3, −5, 6. The line in (C) passes through the origin with direction ratios 5, −2, 3.
So the correct equations are: (A) \(\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}\), (B) \(\frac{x-1}{3}=\frac{y-2}{-5}=\frac{z-3}{6}\), (C) \(\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\), (D) \(2(x-1)+3(y-2)-1(z-3)=0\) and (E) \(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1\).
Checking the statements
(A) Equation of the line passing through the point (1, 2, 3) and parallel to the vector \(3\hat{\imath}+2\hat{\jmath}-2\hat{k}\) is \(\frac{x-1}{3}=\frac{y-2}{2}=\frac{y-3}{-2}\). — Correct. With z in the third ratio (see the note), it is the line through (1, 2, 3) with direction ratios 3, 2, −2.
(B) Equation of line passing through (1, 2, 3) and parallel to the line given by \(\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}\) is \(\frac{x-1}{3}=\frac{y-2}{5}=\frac{z+3}{6}\). — Incorrect. The direction ratios are 3, −5, 6 and the third ratio must be \(\frac{z-3}{6}\).
(C) Equation of line passing through the origin and (5, −2, 3) is \(\frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\). — Correct. Line through (0, 0, 0) with direction ratios 5, −2, 3.
(D) Equation of plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratios 2, 3, −1 is \(2(x-1)+3(y-2)-1(z-3)=0\). — Correct. The line gives the normal 2, 3, −1 of the plane.
(E) Equation of plane with intercepts 2, 3 and 4 on x, y and z-axis respectively is \(2x+3y+4z=1\). — Incorrect. The intercept form is \(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1\); \(2x+3y+4z=1\) meets the x-axis at \(x = \frac{1}{2}\).
Note: It seems that there is a mistake in the original NTA paper, which prints the third ratio of the line in statement (A) as \(\frac{y-3}{-2}\) when it should read \(\frac{z-3}{-2}\). As printed, (A) is not correct, because the point (1, 2, 3) does not satisfy it: y = 2 gives \(\frac{y-2}{2} = 0\) but \(\frac{y-3}{-2} = \frac{1}{2}\). The question can still be solved because (C) and (D) are correct and (B) and (E) are not, and Option (2) is the only option that contains (C) and (D) without (E).