Question

Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.

 

Answer 1

It is given that O is the origin and the coordinates of A are (a, b, c).

The direction ratios of OA are proportional to 

\[a - 0, b - 0, c - 0\]

∴ Direction cosines of OA are

\[\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}\]

The normal vector to the required plane is

`  a  \hat{i }+ b  \hat{j } + c\hat{k } `

The vector equation of the plane through A (a, b, c) and perpendicular to OA is

`  \vec{r} - ( a  \hat{i }+ b   \hat{j }+ c\hat{k}) ] . ( a   \hat{i } + b   \hat{j }+ c\hat{k}) = 0 ( \vec{r} - \vec{a} ) . \vec{n} = 0  `

`   \vec{r} . ( a  \hat{i } + b  \hat{j } + c \hat{k }) = ( a  \hat{i }+ b  \hat{j } + c\hat{k}) . ( a   \hat{i } + b  \hat{j }+ c\hat{k} ) `

`  \vec{r} . ( a  \hat{x } + b  \hat{y }+ c   \hat{z }) = a^2 + b^2 + c^2  `

The Cartesian equation of this plane is

` ( x \hat{i }  + y  \hat{ j } + z   \hat{k }) . ( a   \hat{i }+ b  \hat{j }+ c\hat{k }) = a^2 + b^2 + c^2  `

`  \text{ Or  }ax + by + cz = a^2 + b^2 + c^2 `

Answer 2

It is given that O is the origin and the coordinates of A are (a, b, c).

The direction ratios of OA are proportional to 

\[a - 0, b - 0, c - 0\]

∴ Direction cosines of OA are

\[\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}\]

The normal vector to the required plane is

`  a  \hat{i }+ b  \hat{j } + c\hat{k } `

The vector equation of the plane through A (a, b, c) and perpendicular to OA is

`  \vec{r} - ( a  \hat{i }+ b   \hat{j }+ c\hat{k}) ] . ( a   \hat{i } + b   \hat{j }+ c\hat{k}) = 0 ( \vec{r} - \vec{a} ) . \vec{n} = 0  `

`   \vec{r} . ( a  \hat{i } + b  \hat{j } + c \hat{k }) = ( a  \hat{i }+ b  \hat{j } + c\hat{k}) . ( a   \hat{i } + b  \hat{j }+ c\hat{k} ) `

`  \vec{r} . ( a  \hat{x } + b  \hat{y }+ c   \hat{z }) = a^2 + b^2 + c^2  `

The Cartesian equation of this plane is

` ( x \hat{i }  + y  \hat{ j } + z   \hat{k }) . ( a   \hat{i }+ b  \hat{j }+ c\hat{k }) = a^2 + b^2 + c^2  `

`  \text{ Or  }ax + by + cz = a^2 + b^2 + c^2 `

Answer 3

The equation of the plane in the intercept form is 

\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\]  where a, b and c are the intercepts on the x, y and z-axis, respectively.

It is given that the intercepts made by the plane on the x, y and z-axis are 3, –4 and 2, respectively.

∴ a = 3, b = −4, c = 2

Thus, the equation of the plane is 

\[\frac{x}{3} + \frac{y}{\left( - 4 \right)} + \frac{z}{2} = 1\]

\[ \Rightarrow 4x - 3y + 6z = 12\]

`  Rightarrow ( x   \hat{i }+ y  \hat{ j } + z\hat{k  }) . ( 4   \hat{i } - 3   \hat{j  } + 6 \hat  {k } ) = 12 `

` Rightarrow \vec{r} . ( 4   \hat{i  } - 3  \hat{j  } + 6\hat{ k  } \right) = 12 `

This is the vector form of the equation of the given plane.