The Cartesian equation of a line is frac{x-5} | vedclass

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(B) The given Cartesian equation is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2}$.First,rewrite the equation in the standard form $\frac{x-x_1}{a} =

Vedclass · Accepted Answer

$\vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k})$

Vedclass · Answer

(B) The given Cartesian equation is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{6-z}{2}$.First,rewrite the equation in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.For the third term,$\frac{6-z}{2} = \frac{-(z-6)}{2} = \frac{z-6}{-2}$.So,the equation becomes $\frac{x-5}{3} = \frac{y-(-4)}{7} = \frac{z-6}{-2}$.Comparing this with the standard form,the point $(x_1, y_1, z_1)$ is $(5, -4, 6)$ and the direction ratios $(a, b, c)$ are $(3, 7, -2)$.The position vector of the point is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$ and the direction vector is $\vec{b} = 3\hat{i} + 7\hat{j} - 2\hat{k}$.The vector equation of a line is given by $\vec{r} = \vec{a} + \lambda\vec{b}$.Substituting the values,we get $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} - 2\hat{k})$.Thus,the correct option is $B$.

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{6-z}{2}$. Find the vector equation of the line.

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