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O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of the plane through A at right angle to OA.

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We have the points O (0, 0, 0) and A (a, b, c) where a, b, and c are direction ratios. We need to find the direction cosines of line OA and the equation of the plane through A at right angle to OA.

To begin with,

\vec{OA}=Position\: vector\: of\: A-Position\: vector\: of\:O\\ \Rightarrow \vec{OA}=\left ( a\hat{i}+b\hat{j}+c\hat{k} \right )-\left ( 0\hat{i}+0\hat{j}+0\hat{k} \right ) \\ \Rightarrow \vec{OA}= a\hat{i}-0\hat{i}+b\hat{j}-0\hat{j}+c\hat{k}-0\hat{k}\\ \Rightarrow \vec{OA}= a\hat{i}+b\hat{j}+c\hat{k}

We know, if (a, b, c) are the direction ratios of a given vector, then its direction cosines will be:

\left ( \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} \right )

According to the question, the direction ratios are (a, b, c), therefore the direction cosines of the vector OA are the same as the above formula, that is,

\left ( \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} \right )

Given, the plane is perpendicular to OA. We know, a normal is a line or vector which is perpendicular to a given object. Therefore, we can say:

\vec{n}=\vec{OA}\\ \Rightarrow \vec{n}=a\hat{i}+b\hat{j}+c\hat{k}\\ \left [ \because \vec{OA}=a\hat{i}+b\hat{j}+c\hat{k} \right ]

Also, the vector equation of a plane where the normal is passing   through the plane and passing through is,

\left ( \vec{r}-\vec{a} \right ).\vec{n}=0


\vec{r}-\vec{a} = vector\: from\: \vec{A}\: to\: \vec{R} \\ \vec{a}=Position\: vector\: of\: the\: given\: point\: in\: the\: plane\\ \vec{n}=normal\: vector\: to\: the\: plane

Here, the given point in the plane is A (a, b, c).

\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ \vec{a}=a\hat{i}+b\hat{j}+c\hat{k}\\ \vec{n}=a\hat{i}+b\hat{j}+c\hat{k}\\

Substituting the vectors respectively, we get:

\left (\left (x\hat{i}+y\hat{j}+z\hat{k} \right )-\left (a\hat{i}+b\hat{j}+c\hat{k} \right ) \right ).\left (a\hat{i}+b\hat{j}+c\hat{k} \right )=0\\ \Rightarrow \left (x\hat{i}+y\hat{j}+z\hat{k} -a\hat{i}-b\hat{j}-c\hat{k} \right ).\left (a\hat{i}+b\hat{j}+c\hat{k} \right )=0\\ \Rightarrow \left (\left (x-a \right )\hat{i}+\left (y-b \right )\hat{j}+\left (z-c \right )\hat{k}\right ).\left (a\hat{i}+b\hat{j}+c\hat{k} \right )=0\\ \Rightarrow a\left (x-a \right )+b\left (y-b \right )+c\left (z-c \right )=0\\

Upon simplifying this, we get:

\Rightarrow ax - a^{2} + by-b^{2} + cz - c^{2} =0\\ \Rightarrow ax + by + cz - a^{2}-b^{2}- c^{2} =0\\ \Rightarrow a^{2}+b^{2}+ c^{2} =ax + by + cz

Hence, the required equation of the plane is a² + b² + c² = ax + by + cz.

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