Q

# Give answer! - Three Dimensional Geometry - JEE Main-4

The vector equation of the plane through the line of interectionb of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$ is :

• Option 1)

$\vec{r}\times \left ( \vec{i}-\vec{k} \right )+2=0$

• Option 2)

$\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )-2=0$

• Option 3)

$\vec{r}\times \left ( \vec{i}+\vec{k} \right )+2=0$

• Option 4)

$\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )+2=0$

Views

Equation of plane passing through line of interaction of plane $P_1$ and $P_2$ is

$\left ( x+y+z-1 \right )+\lambda \left ( 2x+3y+4z-5 \right )=0\cdots (i)$

$\Rightarrow \left ( 1+2\lambda \right )x+\left ( 1+3\lambda \right )y+\left ( 1+4\lambda \right )z-1-5\lambda =0\;\;\;\dots(ii)$

plane (ii) and $x-y+z=0$ is perpendicular.

so,

$\left ( 1+2\lambda \right )\times 1+\left ( 1+3\lambda \right )\left ( -1 \right )+\left ( 1+4\lambda \right )\left ( 1 \right )=0$

$1+2\lambda -1-3\lambda +1+4\lambda =0$

$3\lambda +1=0$

$\lambda =-\frac{1}{3}$

Equation of plane is

$\left ( 1-\frac{2}{3} \right )x+\left ( 1-\frac{3}{3} \right )y+\left ( 1+4\left ( -\frac{1}{3} \right ) \right )z-1+\frac{5}{3}=0$

$\frac{1}{3}x+0y-\frac{1}{3}z+\frac{2}{3}=0$

$\Rightarrow x-z+2=0$

vector equation.              $\vec{r}\cdot \left ( \hat{i}-\hat{k}\right )+2=0$

Option 1)

$\vec{r}\times \left ( \vec{i}-\vec{k} \right )+2=0$

Option 2)

$\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )-2=0$

Option 3)

$\vec{r}\times \left ( \vec{i}+\vec{k} \right )+2=0$

Option 4)

$\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )+2=0$

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