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प्रश्न
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
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उत्तर १
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 `
उत्तर २
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 `
उत्तर ३
The equation of the plane in the intercept form 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.
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