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  • State True or False for the given statement: If the foot of the perpendicular drawn from the origin to a plane is (5, -3, -2) then the equation of the plane is r.(5hat{i}-3hat{j}-2hat{k})=38

Q: State True or False for the given statement:

If the foot of the perpendicular drawn from the origin to a plane is (5, -3, -2)  then the equation of the plane is  7r.\left ( 5\hat{i}-3\hat{j}-2\hat{k} \right )=38

Answers (1)

Answer:

Let us take O as the origin, P as the foot of the perpendicular drawn from origin to the plane.

Then the position vector OP is:

\vec{n}=\vec{OP}=5\hat{i}-3\hat{j}-2\hat{k} 

The unit vector of n is: 

\vec{n}=\frac{\vec{n}}{\left | \vec{n} \right |}\\ \hat{n}=\frac{5\hat{i}-3\hat{j}-2\hat{k}}{\sqrt{5^{2}+(-3)^{2}+(-2)^{2}}}=\frac{5}{\sqrt{38}}\hat{i}-\frac{3}{\sqrt{38}}\hat{j}-\frac{2}{\sqrt{4}}\hat{k}

OP =\sqrt{\left ( 5-0 \right )^{2}+\left ( -3-0 \right )^{2}+\left (-2-0 \right )^{2}} \\ =\sqrt{25+9+4}\\ =\sqrt{38}

Now, the equation of the plane with unit normal vector n and having a perpendicular drawn from the origin d is:

\vec{r}.\hat{n}=d

Therefore,

Equation of the given plane will be,

\vec{r}.\left ( \frac{5}{\sqrt{38}}\hat{i}-\frac{3}{\sqrt{38}}\hat{j}-\frac{2}{\sqrt{4}}\hat{k} \right )=\sqrt{38}\\ \Rightarrow \vec{r}.\left ( 5\hat{i}-3\hat{j}-2\hat{k} \right )=38

=> The given statement is True.

Posted by

infoexpert24

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