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Find the distance of the point (2, 12, 5) from the point of intersection of the line $\overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda (3\widehat{i}+4\widehat{j}+2\widehat{k})$and the plane$\overrightarrow{r}.\left( \widehat{i}-2\widehat{j}+\widehat{k} \right)=0.$

Last updated date: 09th Aug 2024
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Hint: First, find out any general point on the given line, $\overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda (3\widehat{i}+4\widehat{j}+2\widehat{k})$. Substitute this point in the given plane and find out the intersection point performing the scalar product of vectors. Use the distance formula to find out the distance accordingly.

We need to find the distance of the point (2, 12, 5) from the point of intersection of the line $\overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda (3\widehat{i}+4\widehat{j}+2\widehat{k})$and the plane $\overrightarrow{r}.\left( \widehat{i}-2\widehat{j}+\widehat{k} \right)=0.$
First, let us consider the given line:
$\overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda (3\widehat{i}+4\widehat{j}+2\widehat{k})$
The above line can be rewritten and expressed as:
$\overrightarrow{r}=\left( 2+3\lambda \right)\widehat{i}+\left( -4+4\lambda \right)\widehat{j}+\left( 2+2\lambda \right)\widehat{k}.........(1)$
Now a general point on this line can be taken as:
$x=2+3\lambda$
$y=-4+4\lambda$
$z=2+2\lambda$
As it was mentioned that the line and the plane intersect each other at a point, then we know that the line equation will satisfy the plane equation at the intersection point.
So, substituting equation (1) in the plane equation, we will have
$\left[ \left( 2+3\lambda \right)\widehat{i}+\left( -4+4\lambda \right)\widehat{j}+\left( 2+2\lambda \right)\widehat{k} \right].\left( \widehat{i}-2\widehat{j}+\widehat{k} \right)=0$
Performing the scalar product or dot product of vectors, we can simplify the above equation as:
$\left( 2+3\lambda \right)1+\left( -4+4\lambda \right)\left( -2 \right)+\left( 2+2\lambda \right)1=0$
$2+3\lambda +8-8\lambda +2+2\lambda =0$
$12-3\lambda =0$
$3\lambda =12$
$\lambda =4$
So, we find the value of $\lambda$ is 4.
Therefore, the coordinates of intersection point will be:
$x=2+3\lambda =14$
$y=-4+4\lambda =12$
$z=2+2\lambda =10$
We know the distance between any two points $\left( a,b,c \right)$ and $\left( d,e,f \right)$ is given by:$\sqrt{{{(d-a)}^{2}}+{{(e-b)}^{2}}+{{(f-c)}^{2}}}$
Now the distance between (2, 12, 5) and (14, 12, 10) is given as:
$\sqrt{{{\left( 14-2 \right)}^{2}}+{{\left( 12-12 \right)}^{2}}+{{\left( 10-5 \right)}^{2}}}$
Solving, further we get:
$\sqrt{{{12}^{2}}+{{0}^{2}}+{{5}^{2}}}$
$\sqrt{169}$
$=13$ units
So, the distance of the point (2, 12, 5) from the point of intersection of the line $\overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda (3\widehat{i}+4\widehat{j}+2\widehat{k})$ and the plane$\overrightarrow{r}.\left( \widehat{i}-2\widehat{j}+\widehat{k} \right)=0$ is 13 units.

Note: While performing the scalar product, make sure that you are multiplying the corresponding components accordingly or you will get a different value of $\lambda$, which further leads to a wrong answer. As distance cannot be expressed in negatives, we have omitted the negative value after solving the square root.