Question

Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.

Answer 1

`  \text{ We know that the vector equation of the plane passing through a point} \vec{a} \text{ and normal to }\vec{n}  \text{ is } `

\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]

` \text{ Substituting }\vec{a} = \text{ 5 }\hat{i }+ \text{ 2 }\hat{j } - \text{ 4 }\hat{k } \text{ and }\vec{n} = \text{ 2 }\hat{i } + \text{ 3 }\hat{j } - \hat{k  }     (\text{ because the direction ratios of  } \vec{n} \text{  are 2, 3, -1)}, \text{ we get } `

` \vec{r} . (\text{ 2 }\hat{i} + \text{ 3 }\hat{  j } - \hat{  k})= \text{ 5 }\hat{  i }+ \text{ 2 }\hat{  j } - \text{ 4 }\hat{  k } \text{ and }\vec{n}. ( 2 \hat{i }+3 \hat{j }- \hat{k } )`

`  ⇒  \vec{r} . (\text{ 2 }\hat{  i } + \text{ 3 }\hat{  j } - \hat{  k}) = 10 + 6 + 4 `

`  ⇒   \vec{r} . (\text{ 2 }\hat{  i } + \text{ 3 }\hat{  j } - \hat{  k}) = 20  `

  ` \text { For Cartesian form, we need to substitute } \vec{r} = \vec{r} . (\text{ x }\hat{  i } + \text{ y }\hat{  j } - \text{ z }\hat{  k}) \text{ in this equation. Then, we get } `

 ` (\text{ x }\hat{i } + \text{ y }\hat{j }  + \text{ z }\hat{k }) .(\text{ 2 }\hat{i } + \text{ 3 }\hat{ j } - \hat{k})= 20  `

\[ \Rightarrow 2x + 3y - z = 20\]