Discussion Forum

If the  distance of points $2\hat{i}+3\hat{j}+\lambda \hat{k}$ from the plane $r.(3\hat{i}+2\hat{j}+6\hat{k})=13$ is  5 units. then $\lambda$=

Answer:

Explanation:

Given point be (2,3,$\lambda$) and equation of the plane be

 $r.(3\hat{i}+2\hat{j}+6\hat{k})=13$

$\therefore$    $(x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+2\hat{j}+6\hat{k})=13$

                                                                    $[\because r.(x\hat{i}+y\hat{j}+z\hat{k})]$

$\Rightarrow$          $3x+2y+6z=13$

or   3x+2y+6z-13=0

 Now, distance of the plane from the point (2,3, $\lambda$)

$d=|\frac{ax_{1}+by_{1}+cz_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}|$

$5=|\frac{3\times2+2\times3+6 \times\lambda-13}{\sqrt{3^{2}+2^{2}+6^{2}}}|$

                                                                          [given , d=5]

 $\pm5=\frac{6+6+6 \times\lambda-13}{\sqrt{9+4+36}}$

$\Rightarrow$     $\pm5=\frac{6\lambda-1}{\sqrt{49}}$

$\therefore$    $5=|\frac{6\lambda-1}{7}|$

 $\Rightarrow$    $\pm35=6\lambda-1$

 $\Rightarrow$      $35=6\lambda-1$

or                     $-35=6\lambda-1$

 $\Rightarrow$    $\lambda=6,\lambda=-\frac{17}{3}$