Find the vector equation of a plane which is | vedclass

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Question

(A) The normal vector is $\vec{n} = 3 \hat{i} + 5 \hat{j} - 6 \hat{k}$.The unit normal vector $\hat{n}$ is given by $\hat{n} = \frac{\vec{n}}{|\vec{n}

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$\vec{r} \cdot \left( \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{70}} \right) = 7$

Vedclass · Answer

(A) The normal vector is $\vec{n} = 3 \hat{i} + 5 \hat{j} - 6 \hat{k}$.The unit normal vector $\hat{n}$ is given by $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{3^2 + 5^2 + (-6)^2}} = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{9 + 25 + 36}} = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{70}}$.The vector equation of a plane at a distance $d$ from the origin with unit normal vector $\hat{n}$ is $\vec{r} \cdot \hat{n} = d$.Substituting the values,we get $\vec{r} \cdot \left( \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{70}} \right) = 7$.

Find the vector equation of a plane which is at a distance of $7$ units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$.

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