1.

Find the vector equation of the plane which is at a distance of \(\frac{7}{\sqrt{38}}\) from the origin and the normal vector from origin is \(2\hat{i}+3\hat{j}-5\hat{k}\)?(a) \(\vec{r}.(\frac{2\hat{i}}{38}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{56}}\)(b) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)(c) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{5\hat{j}}{\sqrt{38}}+\frac{3\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)(d) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{58}}-\frac{3\hat{j}}{\sqrt{37}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)The question was asked in exam.My question is from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12

Answer»

The correct answer is (b) \(\vec{r}.(\frac{2\hat{i}}{\SQRT{38}}+\frac{3\hat{J}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)

The best I can EXPLAIN: Let \(\vec{n}=2\hat{i}+3\hat{j}-5\hat{k}\)

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

HENCE, the required equation of the PLANE is \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)


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