Name: Equation of Plane Making Intercepts on Axes
Uploaded: 2019-07-19
Duration: 3 min 49 s
Description: In 2-D geometry, you have studied the Intercept form of the Equation of the straight line. Could you think of the Intercept form of the equation of a plane?...

Question

Derive the equation of the line in the space passing through a point and parallel to the given vector in both vector form and Cartesian form.

Accepted Answer

Let $l$ be the line which passes through the point $A (x_{1}, y_{1}, z_{1})$ and is parallel to a given vector $\vec{b}$ whose direction ratios are $a, b, c$

Let $\vec{r}$ be any position vector of any arbitrary point $P (x, y, z)$ on the line $l$

Then $\vec{A P}$ is parallel to $\vec{b}$

Therefore, $\vec{A P} = λ \vec{b}$ , $λ$ is some real number

$\Rightarrow \vec{O P} - \vec{O A} = λ \vec{b}$

$\Rightarrow \vec{r} - \vec{a} = λ \vec{b}$

$\Rightarrow \vec{r} = \vec{a} + λ \vec{b}$ is vector form

Now, $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}, \vec{a} = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k} and \vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$

Substituting the values of $\vec{r}, \vec{a} and \vec{b}$ in equation $\vec{r} = \vec{a} + λ \vec{b}$ , then we get:

$x \hat{i} + y \hat{j} + z \hat{k} = x_{1} \hat{i} + y_{1} \hat{j} + z_{1} \hat{k} + λ (a \hat{i} + b \hat{i} + c \hat{k})$

Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$ , we get:

$x = x_{1} + λ a; y = y_{1} + λ b; z = z_{1} + λ c$

On eliminating $λ$ , we get: $\frac{(x - x_{1})}{a} = \frac{(y - y_{1})}{b} = \frac{(z - z_{1})}{c}$ , which is required Cartesian equation.

Derive the equation of the line in the space passing through a point and parallel to the given vector in both vector form and Cartesian form.

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