Test Index

MHTCET 10 May 2024 Shift 2 Paper

Question : 102 of 150

Marks: +1, -0

A(1,2,-1)

2 \hat{i}+\hat{j}-\hat{k}

\hat{i}-\hat{j}+3 \hat{k}

$\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=-9$
$\overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})=9$
$\overline{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})=9$
$\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=-9$

Solution:

Concept:

A plane parallel to two vectors has its normal vector equal to the cross product of those vectors.

The plane equation is

\overline{\mathrm{r}} \cdot \overline{\mathrm{n}} = \overline{\mathrm{a}} \cdot \overline{\mathrm{n}}

, where

\overline{\mathrm{a}}

is a point on the plane.

Explanation:

Given point

A(1,2,-1)

, so

\overline{\mathrm{a}} = \hat{\mathrm{i}} + 2\hat{\mathrm{j}} - \hat{\mathrm{k}}

The direction vectors are

\overline{\mathrm{u}} = 2\hat{\mathrm{i}} + \hat{\mathrm{j}} - \hat{\mathrm{k}}

and

\overline{\mathrm{v}} = \hat{\mathrm{i}} - \hat{\mathrm{j}} + 3\hat{\mathrm{k}}

Normal vector to the plane is:

\overline{\mathrm{n}} = \overline{\mathrm{u}} \times \overline{\mathrm{v}} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -1 \\ 1 & -1 & 3 \end{vmatrix} = 2\hat{\mathrm{i}} - 7\hat{\mathrm{j}} - 3\hat{\mathrm{k}}

Now compute

\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}

(1)(2) + (2)(-7) + (-1)(-3) = 2 - 14 + 3 = -9

Therefore, the vector equation is:

\overline{\mathrm{r}} \cdot (2\hat{\mathrm{i}} - 7\hat{\mathrm{j}} - 3\hat{\mathrm{k}}) = -9

Answer:

Option D:

\overline{\mathrm{r}} \cdot (2\hat{\mathrm{i}} - 7\hat{\mathrm{j}} - 3\hat{\mathrm{k}}) = -9

Go to Question:

Prev Question

Next Question