Q.Obtain the vector equation of the line $\dfrac{x+5}{3} = \dfrac{y+4}{5} = \dfrac{z+5}{6}$.

Vector equation of a line through a point $A(\bar a)$ parallel to $\bar b$ is $\bar r=\bar a+\lambda\bar b$.

The line passes through the point with position vector $\bar a = 4\hat i-\hat j+2\hat k$ and is parallel to $\bar b=-2\hat i-\hat j+\hat k$.

The vector equation of a line through a fixed point $\bar a$ and parallel to $\bar b$ is

$\bar r=\bar a+\lambda \bar b$

Substituting,

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

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Vector equation: (point) + $\lambda$(direction).

Line through position vector $\bar a=4\hat i-\hat j+2\hat k$, parallel to $\bar b=-2\hat i-\hat j+\hat k$.

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Vector equation of a line through a point $\bar a$ in the direction $\bar b$ is $\bar r = \bar a + \lambda \bar b$.

The line passes through the point with position vector $\bar a = 4\hat i - \hat j + 2\hat k$ and has direction vector $\bar b = -2\hat i + \hat j + \hat k$.

The vector equation of a line through a given point with position vector $\bar a$ along direction $\bar b$ is: …