The Cartesian equation of a line is frac{x+3} | vedclass

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Question

(A) Comparing the given Cartesian equation $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$ with the standard form $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\fr

Vedclass · Accepted Answer

$\vec{r}=(-3 \hat{i}+5 \hat{j}-6 \hat{k})+\lambda(2 \hat{i}+4 \hat{j}+2 \hat{k})$

Vedclass · Answer

(A) Comparing the given Cartesian equation $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$ with the standard form $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$,we get:$x_{1}=-3, y_{1}=5, z_{1}=-6$ and $a=2, b=4, c=2$.Thus,the line passes through the point $A(-3, 5, -6)$ and is parallel to the vector $\vec{b} = 2\hat{i} + 4\hat{j} + 2\hat{k}$.The position vector of the point is $\vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}$.The vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.Substituting the values,we get $\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})$.

The Cartesian equation of a line is $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$. Find the vector equation for the line.

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