Find the vector equation of the plane which is at a distance of \(\frac{7}{\sqrt{38}}\) from the origin and the normal v
Posted: Wed Jul 13, 2022 8:09 pm
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) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{58}}-\frac{3\hat{j}}{\sqrt{37}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)
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) \(\vec{r}.(\frac{2\hat{i}}{\sqrt{58}}-\frac{3\hat{j}}{\sqrt{37}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}\)