Q

# Solve this problem - Three Dimensional Geometry - JEE Main-5

The length of the perpendicular drawn from the point (2,1,4) to the plane

containing the lines $\vec{r}=(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})$  and

$\vec{r}=(\hat{i}+\hat{j})+\mu (-\hat{i}+\hat{j}-2\hat{k})$ is :

• Option 1)

3

• Option 2)

$\frac{1}{3}$

• Option 3)

$\sqrt{3}$

• Option 4)

$\frac{1}{\sqrt{3}}$

Views

Perpendicula vector to the plane

$\vec{w}=$ $\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 1& 2 &-1 \\ -1&1 &-2 \end{vmatrix}=-3\hat{i}+3\hat{j}+3\hat{k}$

Equation of the plane

$-3(x-1)+3(y-1)+3z=0$

Equation => $x-y-z=0$

$d_{(2,1,4)}=\frac{|2-1-4|}{\sqrt{1^{2}+1^{2}+1^{2}}}=\sqrt3$

Option 1)

3

Option 2)

$\frac{1}{3}$

Option 3)

$\sqrt{3}$

Option 4)

$\frac{1}{\sqrt{3}}$

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