The Cartesian equation of a line is frac{x+2} | vedclass

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Question

(A) The Cartesian equation of a line is given by $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.Comparing this with the given equation $\frac{x-(-2

Vedclass · Accepted Answer

$\bar{r}=(-2 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+5 \hat{k})$

Vedclass · Answer

(A) The Cartesian equation of a line is given by $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.Comparing this with the given equation $\frac{x-(-2)}{3}=\frac{y-4}{2}=\frac{z-5}{5}$,we identify the point $(x_1, y_1, z_1) = (-2, 4, 5)$ through which the line passes.The direction ratios $(a, b, c)$ are $(3, 2, 5)$.The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.Here,$\vec{a} = -2 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\vec{b} = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$.Thus,the vector equation is $\vec{r} = (-2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} + 5 \hat{k})$.

The Cartesian equation of a line is $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$,then the vector equation of the line is

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