Find the vector equation of the line frac{x - | vedclass

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(C) The given equation of the line is $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.We can rewrite this in standard form as $\frac{x - 2}{2} = \frac{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given equation of the line is $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.We can rewrite this in standard form as $\frac{x - 2}{2} = \frac{2(y - 5/2)}{-3} = \frac{z + 1}{0}$.This simplifies to $\frac{x - 2}{2} = \frac{y - 5/2}{-3/2} = \frac{z + 1}{0}$.This shows that the line passes through the point $(2, 5/2, -1)$ and has direction ratios proportional to $(2, -3/2, 0)$.The position vector of the point is $\vec{a} = 2\hat{i} + \frac{5}{2}\hat{j} - \hat{k}$.The vector parallel to the line is $\vec{b} = 2\hat{i} - \frac{3}{2}\hat{j} + 0\hat{k}$.The vector equation of a line is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.Substituting the values,we get $\vec{r} = (2\hat{i} + \frac{5}{2}\hat{j} - \hat{k}) + \lambda (2\hat{i} - \frac{3}{2}\hat{j} + 0\hat{k})$.

Find the vector equation of the line $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.

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