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

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(A) The Cartesian equation of the line is given by $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$ The general form of the Cartesian equation of a line

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

(A) The Cartesian equation of the line is given by $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$ The general form of the Cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} .$ Comparing the given equation with the general form,we get the point $(x_1, y_1, z_1) = (5, -4, 6)$ and direction ratios $(a, b, c) = (3, 7, 2) .$ The position vector of the point is $\vec{a} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k} .$ The vector parallel to the line 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 $\vec{r} = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + \lambda(3 \hat{i} + 7 \hat{j} + 2 \hat{k}) ,$ where $\lambda$ is a scalar.

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

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