The vector equation of the plane which is at a distance of 6 29 from the origin and its normal vector from the origin is 2 i - 3 j + 4 k is:

Question

The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

Smart Achievers · Accepted Answer

(2)

Ans. $\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$

Explanation:

Let $\vec{n} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} .$ Then,

$\hat{n} = \frac{\vec{n}}{| \vec{n} |} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{4 + 9 + 16}} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}}$

Hence, the required equation of the plane is

$\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$

Solution

Q.
The vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ is:

1 $\vec{r} \cdot (2 \hat{i} - 3 \hat{j} + 4 \hat{k}) = 6$

2 $\vec{r} \cdot (\frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k}) = \frac{6}{\sqrt{29}}$

3 Both (a) and (b)

4 None of the above

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Solution

Q. The vector equation of the plane which is at a distance of 629 from the origin and its normal vector from the origin is 2i^-3j^+4k^ is:

1 r→·(2i^-3j^+4k^)=6

2 r→·(229i^-329j^+429k^)=629