Question

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 $\overrightarrow{r}.(\widehat{i}-2\widehat{j}+\widehat{k})=0.$

Answer 1

Hint: In the above question, first of all, we will take a general point on the line and then we will find the point of intersection of the line and the plane by substituting the general point of the line in the equation of the plane and then we will find the distance between the points. Also, we will use the distance formula to find the distance between two given points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ as $\text{distance}=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$.
Complete step by step answer:
We have a line $\overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda (3\widehat{i}+4\widehat{j}+2\widehat{k})$ and a plane $\overrightarrow{r}.(\widehat{i}-2\widehat{j}+\widehat{k})=0$ given in the question. To find the point of intersection of these two, first, we have to find a general point on the line. So, we can rewrite the equation of line as $\overrightarrow{r}=\left( 2+3\lambda \right)\widehat{i}+\left( -4+4\lambda \right)\widehat{j}+\left( 2+2\lambda \right)\widehat{k}$. Now, we can write $\overset{\to }{\mathop{r}}\,=x\widehat{i}+y\widehat{j}+z\widehat{k}$ to get the general point. So, we get $x\widehat{i}+y\widehat{j}+z\widehat{k}=\left( 2+3\lambda \right)\widehat{i}+\left( -4+4\lambda \right)\widehat{j}+\left( 2+2\lambda \right)\widehat{k}$. Now, equating x, y and z terms, we get the coordinates of the general point as, $x=2+3\lambda ,y=-4+4\lambda ,z=2+2\lambda$.
Since we know that the line and plane intersect, the general point that we have found would satisfy the equation of the plane. So, we can substitute it to find the value of $\lambda$.

& \left[ \left( 2+3\lambda \right)\widehat{i}+\left( -4+4\lambda \right)\widehat{j}+\left( 2+2\lambda \right)\widehat{k} \right].(\widehat{i}-2\widehat{j}+\widehat{k})=0 \\\ & \left[ \left( 2+3\lambda \right).1+\left( -4+4\lambda \right).(-2)+\left( 2+2\lambda \right).1 \right]=0 \\\ & 12-3\lambda =0 \\\ & \lambda =4 \\\ \end{aligned}$$ Now, we can substitute the value $$\lambda =4$$ to find the point of intersection. So, we will get: (2+3(4), -4 +4(4), 2+2(4)) = (14, 12, 10). We can find the distance of this point (14, 12, 10) from (2,12, 5) using the distance formula between two points $$\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$$ given by $$\text{distance}=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$$. So, substituting the points, we get $$\begin{aligned} & =\sqrt{{{(14-2)}^{2}}+{{(12-12)}^{2}}+{{(10-5)}^{2}}} \\\ & =\sqrt{{{12}^{2}}+{{5}^{2}}} \\\ & =13 \\\ \end{aligned}$$ Therefore, the distance between the given point and the point of intersection between the line and the plane is 13. Note: Be careful while doing calculation as there is a chance that you might make a mistake and you will get the incorrect answer. Also, remember the formula to calculate the distance between two given points and the way to find the point of intersection between a line and a plane. A very common mistake is while applying the distance formula, you might use $$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}-{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}-{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$$ instead of $$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$$.