Q.The vector equation of the line $\dfrac{x-5}{7} = \dfrac{y+4}{7} = \dfrac{6-z}{2}$ is ..........

Rewrite the Cartesian equation so all terms have the form $\dfrac{\text{coordinate}-\text{point}}{\text{direction ratio}}$, then read off the point and direction vector.

Given: $\dfrac{x-5}{7}=\dfrac{y+4}{7}=\dfrac{6-z}{2}$

Rewrite the third term: $\dfrac{6-z}{2}=\dfrac{-(z-6)}{2}=\dfrac{z-6}{-2}$

So: $\dfrac{x-5}{7}=\dfrac{y-(-4)}{7}=\dfrac{z-6}{-2}$

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The Cartesian symmetric form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ directly gives a point $(x_1,y_1,z_1)$ on the line and direction ratios $\langle a,b,c\rangle$, from which the vector equation follows.

The given line is:

$\frac{x-5}{-4} = \frac{y-3}{5} = \frac{z+3}{-8}$

Comparing with $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$, the point on the line is $(5,3,-3)$ and direction ratios are $\langle -4,5,-8\rangle$.

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Cartesian equation of a line through a point with given direction cosines.

A line through $(x_1,y_1,z_1)$ with direction cosines $l,m,n$ has Cartesian equation

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The line passes through $(x_1,y_1,z_1)$ where each numerator vanishes: $(11,12,-13)$.

Write $\frac{x-11}{12}=\frac{y-12}{13}=\frac{z+13}{14}$. Setting each fraction to $0$ gives $x=11$, $y=12$, …

Read the fixed point and direction ratios straight off the Cartesian form and assemble the vector equation r = a + λb.

The Cartesian equation $\dfrac{x-1}{2}=\dfrac{y+2}{3}=\dfrac{z-5}{-1}$ is of the standard form $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$, which passes through $(x_1,y_1,z_1)=(1,-2,5)$ with direction ratios $(a,b,c)=(2,3,-1)$.

The corresponding vector equation is:

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Read the point and direction ratios straight off the Cartesian symmetric form, then write the vector equation $\vec r = \vec a + \mu \vec b$.

The Cartesian equation is $\dfrac{x-5}{-4} = \dfrac{y-3}{5} = \dfrac{z-(-3)}{-8}$.

Comparing with $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$: the line passes through the point $(5, 3, -3)$ and has direction ratios $(-4, 5, -8)$.

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The point $(x_0,y_0,z_0)$ in $\dfrac{x-x_0}{a}=\cdots$ lies on the line, here $(100,99,98)$.

The symmetric form $\dfrac{x-100}{101}=\dfrac{y-99}{102}=\dfrac{z-98}{103}$ passes through the point $(100,99,98)$, obtained by setting each fraction equal to $0$.

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A cartesian line $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ passes through $(x_0,y_0,z_0)$ with direction $(a,b,c)$; write this as a vector equation.

Given $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$, the line passes through the point $(5,-4,6)$ with direction ratios $(3,7,2)$.

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Rewrite each fraction in the form (variable − constant)/(direction ratio) to read off the point the line passes through.

$\dfrac{2x-3}2 = \dfrac{2y+3}{-5} = \dfrac z2$

Rewrite each term: $\dfrac{2x-3}2 = \dfrac{2\left(x-\frac32\right)}2 = x-\dfrac32 = \dfrac{x-\frac32}1$

$\dfrac{2y+3}{-5} = \dfrac{2\left(y+\frac32\right)}{-5} = \dfrac{y-\left(-\frac32\right)}{-5/2}$

$\dfrac z2 = \dfrac{z-0}2$ …

read off the point and direction ratios from the cartesian form

$\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ passes through $(1,2,3)$ with direction ratios $(2,3,4)$.

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