Question:
An equation of the plane passing through the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and passing through (1, 1, 1) is
An equation of the plane passing through the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and passing through (1, 1, 1) is
Updated On: Apr 26, 2024
- 2x + 3y + 4z = 9
- x + y + z = 3
- x + 2y + 3z = 6
- 20x + 23y + 26z = 69
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The Correct Option is D
Solution and Explanation
The equation of the plane through the line of intersection of the given planes is (x + y + z - 6) + $\lambda$ (2x + 3y + 4z + 5) = 0 ... (1)
If equation (1) passes through (1, 1, 1), we have
$-3+14 \lambda \, = 0 \, \, \Rightarrow \, \lambda = \frac{3}{14}$
Putting $\lambda \, =\frac{3}{14}$ in (1), we obtain the
equation of the required plane as
$(x+y+z-6) \, + \, \frac{3}{14}$(2x+3y+4z+5)=0
$\Rightarrow$ 20x+23y+26z-69=0
If equation (1) passes through (1, 1, 1), we have
$-3+14 \lambda \, = 0 \, \, \Rightarrow \, \lambda = \frac{3}{14}$
Putting $\lambda \, =\frac{3}{14}$ in (1), we obtain the
equation of the required plane as
$(x+y+z-6) \, + \, \frac{3}{14}$(2x+3y+4z+5)=0
$\Rightarrow$ 20x+23y+26z-69=0
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Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)