The vector equation of the plane r→=(i^−j^)+λ(i^+j^+k^)+μ(i^−2j^+3k^) in scalar product form, is

Clearly, given plane passes through the point a=i^−j^ and and is parallel to the vectors b→=i^+j^+k^ and c→=i^−2j^+3k^So, it is perpendicular to the vector n→ given byn→=b→×c→=i^ j^ k^1 1 11 −2 3=5i^−2j^−3k^The vector equation of the plane in scalar product form isr→⋅n→=a→⋅n→⇒ r→⋅(5i^−2j^−3k^)=(i^−j^)⋅(5i^−2j^−3k^)⇒ r→⋅(5i^−2j^−3k^)=7

The vector equation of the plane r→=(i^−j^)+λ(i^+j^+k^)+μ(i^−2j^+3k^) in scalar product form, is

Clearly, given plane passes through the point a=i^−j^ and and is parallel to the vectors b→=i^+j^+k^ and c→=i^−2j^+3k^So, it is perpendicular to the vector n→ given byn→=b→×c→=i^ j^ k^1 1 11 −2 3=5i^−2j^−3k^The vector equation of the plane in scalar product form isr→⋅n→=a→⋅n→⇒ r→⋅(5i^−2j^−3k^)=(i^−j^)⋅(5i^−2j^−3k^)⇒ r→⋅(5i^−2j^−3k^)=7

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The vector equation of the plane $\vec{r} = (\hat{i} - \hat{j}) + λ (\hat{i} + \hat{j} + \hat{k}) + μ (\hat{i} - 2 \hat{j} + 3 \hat{k})$ in scalar product form, is

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$\vec{r} \cdot (5 \hat{i} + 2 \hat{j} + 3 \hat{k}) = 7$

none of these

$\vec{r} \cdot (5 \hat{i} - 2 \hat{j} - 3 \hat{k}) = 7$

$\vec{r} \cdot (5 \hat{i} - 2 \hat{j} - 3 \hat{k}) = - 7$

answer is A.

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Detailed Solution

Clearly, given plane passes through the point $a = \hat{i} - \hat{j}$ and and is parallel to the vectors $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - 2 \hat{j} + 3 \hat{k}$

So, it is perpendicular to the vector $\vec{n}$ given by

$\vec{n} = \vec{b} \times \vec{c} = |\begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 1 1 1 \\ 1 - 2 3 \end{array}| = 5 \hat{i} - 2 \hat{j} - 3 \hat{k}$

The vector equation of the plane in scalar product form is

$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$ $\Rightarrow \vec{r} \cdot (5 \hat{i} - 2 \hat{j} - 3 \hat{k}) = (\hat{i} - \hat{j}) \cdot (5 \hat{i} - 2 \hat{j} - 3 \hat{k})$