If vector equation of the plane x - 2 2 = 2 y - 5 - 3 = z + 1 , is r = ( 2 i + 5 2 j - k ) + ( 2 i - 3 2 j + p k ) , then p is equal to:

Question

If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to:

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Explanation:

The given line is

$\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$

$\Rightarrow$ $\frac{x - 2}{2} = \frac{y - \frac{5}{2}}{- \frac{3}{2}} = \frac{z + 1}{0}$

This shows that the given line passes through the point $(2, \frac{5}{2}, - 1)$ and has direction ratios $(2, - \frac{3}{2}, 0) .$ Thus, given line passes through the point having position vector $\vec{a} = 2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}$ and is parallel to the vector $\vec{b} = (2 \hat{i} - \frac{3}{2} \hat{j} - 0 \hat{k}) .$

So, its vector equation is

$\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} - 0 \hat{k})$

Hence, $p = 0$

Solution

Q.
If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to:

1 0

2 1

3 2

4 3

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Solution

Q. If vector equation of the plane x-22=2y-5-3=z+1, is r→=(2i^+52j^-k^)+λ(2i^-32j^+pk^), then p is equal to:

1 0

2 1

3 2

4 3

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Q.
If vector equation of the plane $\frac{x - 2}{2} = \frac{2 y - 5}{- 3} = z + 1,$ is $\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})$ , then $p$ is equal to: