The vector equation of a plane passing through a point having position vector 2i^+3j^−4k^ and perpendicular to the vector 2i^−j^+2k^, is

We know that the vector equation of a plane passing through a point a → and normal to n→ is(r→−a→)⋅n→=0 or, r→⋅n→=a→⋅n→Here, a→=2i^+3j^−4k^ and n→=2i^−j^+2k^ So, the equation of the required plane is r→⋅(2i^−j^+2k^)=(2i^+3j^−4k^)⋅(2i^−j^+2k^)⇒r→⋅(2i^−j^+2k^)=−7

The vector equation of a plane passing through a point having position vector 2i^+3j^−4k^ and perpendicular to the vector 2i^−j^+2k^, is

We know that the vector equation of a plane passing through a point a → and normal to n→ is(r→−a→)⋅n→=0 or, r→⋅n→=a→⋅n→Here, a→=2i^+3j^−4k^ and n→=2i^−j^+2k^ So, the equation of the required plane is r→⋅(2i^−j^+2k^)=(2i^+3j^−4k^)⋅(2i^−j^+2k^)⇒r→⋅(2i^−j^+2k^)=−7

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The vector equation of a plane passing through a point having position vector $2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and perpendicular to the vector $2 \hat{i} - \hat{j} + 2 \hat{k}$ , is

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none of these

$\vec{r} \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) = - 7$

$\vec{r} \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) = 4$

$\vec{r} \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) = 7$

answer is B.

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Detailed Solution

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

$(\vec{r} - \vec{a}) \cdot \vec{n} = 0 or, \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$

Here, $\vec{a} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ and $\vec{n} = 2 \hat{i} - \hat{j} + 2 \hat{k}$

So, the equation of the required plane is

$\begin{array}{l} \vec{r} \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) = (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) \\ \Rightarrow \vec{r} \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) = - 7 \end{array}$