Question

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 1

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 \(\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)