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The plane — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThe plane1 Mark
Question
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:
  1. $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
  2. $\vec{\text{r}}.(\hat{\text{i}}-3\hat{\text{k}})=10$
  3. $\vec{\text{r}}.(3\hat{\text{i}}+\hat{\text{k}})=10$
  4. $\text{None of these}$
✓

Answer

  1. $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$

Solution:

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

$\begin{vmatrix}\text{x}+2&\text{y}+3&\text{z}-4\\3&-2&-1\\3&5&-1\end{vmatrix}=0$

$\Rightarrow7(\text{x}+2)+0(\text{y}+3)+21(\text{y}-4)=0$

$\Rightarrow\text{x}+2+3\text{z}-12=0$

$\Rightarrow\text{x}+3\text{z}=10$ or $\vec{\text{r}}.\big(\hat{\text{i}}+3\hat{\text{k}}\big)=10$

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