A. Equation of the line passing through the point (1, 2, 3) and parallel to the vector $3\hat i+2\hat j-2\hat k$ is $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}$ 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 $\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{6}$ C. Equation of line passing through the origin and (5, -2, 3) is 2-3 D. Equation of plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratio's 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: |
A only A, C, D only C, D, E only D,E only |
A only |
The correct answer is option (1) → A only (A) Equation of line will be $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}$ (Correct Statement) (B) Given line → $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}$ $⇒ \frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}$ direction ratios → 3, -5, 6 So line passing through (1, 2, 3) $\frac{x-1}{3}=\frac{y-2}{-5}=\frac{z-3}{6}$ (Incorrect statement) (C) Vague statement "2 - 3" is not form of a line (Incorrect statement) (D) Eq. of plane through (1, 2, 3) and perpendiculat to (2, 3, -1) vector is $2(x-1)+3(y-2)-1(x-3)=0$ (Incorrect statement) (E) Equation of plane with intercepts 2, 3, 4 is $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ (Incorrect statement) ⇒ Only statement A is correct. |