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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry1 Mark
MCQ
The vector equation of the plane containing the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$ and the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ is:
  • ✓
    $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
  • B
    $\vec{\text{r}}.(\hat{\text{i}}-3\hat{\text{k}})=10$
  • C
    $\vec{\text{r}}.(3\hat{\text{i}}+\hat{\text{k}})=10$
  • D
    None of these
✓

Answer

Correct option: A.
$\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
Let the direction ratio of the required plane be proportinal to $a, b, c.$
Scince the required plane contains the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
It must pass through the point $(-2, -3, 4)$ and it should be parallel to the line.
So, the equation of the plane is
$a(x + 2) + b(y + 3) + c(z - 4) = 0 ....(1)$ and
$3a - 2b - c = 0 ....(2)$
It is given that plane $(1)$ passes through the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ or $(1, 2, 3).$
$a(1 + 2) + b(2 + 3) + c(3 - 4) = 0$
$3a + 5b - c = 0 .......(3)$
So,
Solving $(1) (2)$ and $(3),$ we get
$\begin{vmatrix}\text{x}+2 \text{y}+3 \text{z}-43 -2 -13 5 -1\end{vmatrix}=0$
$\Rightarrow7(\text{x}+2)+0(\text{y}+3)+21(\text{y}-4)=0$
$\Rightarrow\text{x}+2+3\text{z}-12=0$
$\Rightarrow\text{x}+3\text{z}=10$ or $\vec{\text{r}}.\big(\hat{\text{i}}+3\hat{\text{k}}\big)=10$

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