Q101 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

The plane \(3 x+4 y+6 z+7=0\) is rotated about the line \(\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})\) until the plane passes through origin. The equation of the plane in the new position is

  • Correct answer
    A
    \(x+y+z=0\)
  • B
    \(6 x+3 y-4 z=0\)
  • C
    \(4 x-5 y-2 z=0\)
  • D
    \(x+2 y+4 z=0\)

Solution

Equation of plane passes through the origin and containing the line \(\begin{aligned} & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \text { is } \\ & (\mathbf{r}-0) \cdot[(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \times(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}})]=0 \\ & \Rightarrow \mathbf{r} \cdot[\hat{\mathbf{i}}(2-9)-\hat{\mathbf{j}}(1+6)+\hat{\mathbf{k}}(-3-4)]=0 \\ & \Rightarrow \mathbf{r} \cdot(-7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-7 \hat{\mathbf{k}})=0 \\ & \Rightarrow \mathbf{r} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \text { or } x+y+z=0 \end{aligned}\) Hence, option (1) is correct.

Hence, the correct answer is option: 'A'.

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