​ Find the vector equation of the plane passing through the point (3i + 4j + 2k) and parallel to the vectors (i + 2j + 3k) and (i - j + k) ​

Question

Find the vector equation of the plane passing through the point (3\(\hat{i}\) + 4\(\hat{j}\) + 2\(\hat{k}\)) and parallel to the vectors (\(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\)) and (\(\hat{i}\) - \(\hat{j}\) + \(\hat{k}\))

Answer 1

Given \(\bar{r}\) = (\(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\)) and \(\bar{r'}\) = \(\hat{i}\) - \(\hat{j}\) + \(\hat{k}\) are two lines to which a plane is parallel and it passes through the point 3\(\hat{i}\) + 4\(\hat{j}\) + 2\(\hat{k}\)

To find – The equation of the plane 

Tip – A plane parallel to two vectors will have its normal in a direction perpendicular to both the vectors, which can be evaluated by taking their cross product

The equation of the plane maybe represented as 5x + 2y - 3z + d = 0 

Now, this plane passes through the point (3, 4, 2) 

Hence, 

5 × 3 + 2 × 4 – 3 × 2 + d = 0 

⇒ d = - 17 

The Cartesian equation of the plane : 5x + 2y - 3z - 17 

= 0 i.e. 5x + 2y - 3z = 17 

The vector equation : \(\bar{r'}\). (5\(\hat{i}\) + 2\(\hat{j}\) - 3\(\hat{k}\)) = 17