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
- 2x + 3y + 4z = 9
- x + y + z = 3
- x + 2y + 3z = 6
- 20x + 23y + 26z = 69
The Correct Option is D
Solution and Explanation
We are required to find the equation of a plane passing through the line of intersection of two planes and a point \((1,1,1)\).
Given planes:
\(P_1: x + y + z = 6\)
\(P_2: 2x + 3y + 4z + 5 = 0\)
Any plane passing through their line of intersection is written as:
\((x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0\)
Expanding:
\((1 + 2\lambda)x + (1 + 3\lambda)y + (1 + 4\lambda)z + (5\lambda - 6) = 0\)
Since the plane passes through \((1,1,1)\), substitute these values:
\((1 + 2\lambda) + (1 + 3\lambda) + (1 + 4\lambda) + (5\lambda - 6) = 0\)
Simplify:
\(3 + 14\lambda - 6 = 0\)
\(14\lambda - 3 = 0\)
\(\lambda = \frac{3}{14}\)
Substitute back into the plane equation:
\((1 + 2\cdot \frac{3}{14})x + (1 + 3\cdot \frac{3}{14})y + (1 + 4\cdot \frac{3}{14})z + (5\cdot \frac{3}{14} - 6) = 0 \)
Simplifying coefficients:
\(\frac{20}{14}x + \frac{23}{14}y + \frac{26}{14}z - \frac{69}{14} = 0\)
Multiplying throughout by 14:
\(20x + 23y + 26z = 69\)
Final answer: \(20x + 23y + 26z = 69\)
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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\(\hat{k}\)