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VITEEE 2019 Paper

Section: Mathematics

Question : 92 of 125

Marks: +1, -0

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

\vec{r} = \left(2\hat{i} + \frac{5}{2}\hat{j} - \hat{k}\right)

+ \lambda \left(2\hat{i} - \frac{3}{2}\hat{j} + p\hat{k}\right)

then p is equal to

Solution:

The given line is

\frac{x-2}{2} = \frac{2y-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

\left(2, \frac{5}{2}, -1\right)

and has direction ratios

\left(2, -\frac{3}{2}, 0\right)

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} = \left(2\hat{i} + \frac{3}{2}\hat{j} - 0\hat{k}\right)

So, its vector equation is

\vec{r} = \left(2\hat{i} + \frac{5}{2}\hat{j} - \hat{k}\right)

+ \lambda \left(2\hat{i} - \frac{3}{2}\hat{j} + 0\hat{k}\right)

Hence, p = 0

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