Question

How do I find the Cartesian equation of a plane containing two given lines?

Answer 1

Given the equations of lines. We have to find the Cartesian equation of the plane that will contain the given lines. The required plane contains the lines which are parallel and non- coincident.

Complete step by step answer:
To find the equation of the plane, first compare the two equations with the general form of vector equation of the line $\vec r = \vec a + \lambda \vec b$.

Then, determine the value of $\overrightarrow {{a_1}}$, $\overrightarrow {{a_2}}$, $\overrightarrow {{b_1}}$ and $\overrightarrow {{b_2}}$

The equation of the plane containing two given lines $\overrightarrow {{r_1}} = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}}$ and $\overrightarrow {{r_2}} = \overrightarrow {{a_2}} + \lambda \overrightarrow {{b_2}}$ must pass through $\overrightarrow {{a_1}}$. Also, the plane must have $\overrightarrow {{b_1}}$ and $\overrightarrow {{b_2}}$ parallel to it.

Then, determine the normal vector to the required plane using the formula, $\overrightarrow n = \overrightarrow {{b_1}} \times \overrightarrow {{b_2}}$

Then, the normal vector is obtained.

Then, we will obtain the vector equation of the plane which is given by:

$\left( {\vec r - \vec a} \right)\vec n = \vec 0$

$\vec r \cdot \vec n = \overrightarrow {{a_1}} \cdot \vec n$

Then, substitute the value of normal vector to obtain the equation of plane in vector form, $\vec r\left( {a\hat i + b\hat j + c\hat k} \right) = d$.

Then, we will rewrite the equation in Cartesian form, $ax + by + cz = d$

For example, consider the equation of given lines in vector form,

$\vec r = \left( {2\hat i + \hat j - 3\hat k} \right) + {\lambda _1}\left( {\hat i + 2\hat j + 5\hat k} \right)$ ……(1)

$\vec r = \left( {3\hat i + 3\hat j + 2\hat k} \right) + {\lambda _2}\left( {3\hat i - 2\hat j + 5\hat k} \right)$ ……(2)

Now, we will compare the equation (1) and (2) by standard form of vector equation $\vec r = \vec a + \lambda \vec b$

$\Rightarrow \overrightarrow {{a_1}} = 2\hat i + \hat j - 3\hat k$

$\Rightarrow \overrightarrow {{a_2}} = 3\hat i + 3\hat j + 2\hat k$

$\Rightarrow \overrightarrow {{b_1}} = \hat i + 2\hat j + 5\hat k$

$\Rightarrow \overrightarrow {{b_2}} = 3\hat i - 2\hat j + 5\hat k$

Now, the equation of plane will pass through $\overrightarrow {{a_1}} = 2\hat i + \hat j - 3\hat k$

Now, determine the equation of normal vector using the formula $\overrightarrow n = \overrightarrow {{b_1}} \times \overrightarrow {{b_2}}$

\Rightarrow \overrightarrow n = \left[ {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 1&2&5 \\\ 3&{ - 2}&5 \end{array}} \right]

$\Rightarrow \overrightarrow n = \hat i\left( {10 + 10} \right) - \hat j\left( {5 - 15} \right) + \hat k\left( { - 2 - 6} \right)$

On simplifying the expression, we get:

$\Rightarrow \overrightarrow n = 20\hat i + 10\hat j - 8\hat k$

Now, we will find the vector equation of the plane by substituting the value of normal vector into the formula $\vec r \cdot \vec n = \overrightarrow {{a_1}} \cdot \vec n$

$\Rightarrow \vec r \cdot \left( {20\hat i + 10\hat j - 8\hat k} \right) = \left( {2\hat i + \hat j - 3\hat k} \right) \cdot \left( {20\hat i + 10\hat j - 8\hat k} \right)$

Apply the dot product of two vectors.

$\Rightarrow \vec r \cdot \left( {20\hat i + 10\hat j - 8\hat k} \right) = 2 \times 20 + 1 \times 10 + \left( { - 3} \right) \times \left( { - 8} \right)$

$\Rightarrow \vec r \cdot \left( {20\hat i + 10\hat j - 8\hat k} \right) = 40 + 10 + 24$

$\Rightarrow \vec r \cdot \left( {20\hat i + 10\hat j - 8\hat k} \right) = 74$

Divide both sides of equation by 2.

$\Rightarrow \vec r \cdot \left( {10\hat i + 5\hat j - 4\hat k} \right) = 37$

Therefore, the vector equation of the plane is $\vec r \cdot \left( {10\hat i + 5\hat j - 4\hat k} \right) = 37$
Now, we will rewrite the equation in Cartesian form.

$10x + 5y - 4z = 37$

Note: Please note that this is necessary to check whether the equation of lines are parallel or perpendicular to each other before solving such types of questions.