Question

The equation of the plane passing through the intersection of the planes $x+2y-z=3$ and $x+y-3z=5$ and passing through the point $(1,-1,0)$ is

• A. $x+7y+6z+6=0$
• B. $x-6y-7z+5=0$
• C. $x+7y+6z+5=0$
• D. $x+6y-7z-5=0$
• E. $x+6y+7z+5=0$

Hint:

Algebra Tip: To avoid fractions, you can also write the family of planes as $k_1 P_1 + k_2 P_2 = 0$. Using the point gives $-4k_1 - 5k_2 = 0 \implies 4k_1 = -5k_2$. Let $k_1 = 5$ and $k_2 = -4$. This skips fraction arithmetic!

Answer 1

Answer: (E)

Concept:
The equation of any plane passing through the intersection line of two planes $P_1 = 0$ and $P_2 = 0$ belongs to the family of planes defined by $P_1 + \lambda P_2 = 0$. By substituting the coordinates of the known third point into this family equation, we can solve for the specific scalar $\lambda$.

Step 1: Set up the family of planes equation.
First, rewrite the plane equations so they equal zero: $P_1: x + 2y - z - 3 = 0$ $P_2: x + y - 3z - 5 = 0$ The combined equation is: $$(x + 2y - z - 3) + \lambda(x + y - 3z - 5) = 0$$

Step 2: Substitute the given point to solve for $\lambda$.
The target plane passes through $(x, y, z) = (1, -1, 0)$. Plug this point in: $$(1 + 2(-1) - 0 - 3) + \lambda(1 + (-1) - 3(0) - 5) = 0$$

Step 3: Simplify and isolate $\lambda$.
Simplify the brackets: $$(1 - 2 - 3) + \lambda(1 - 1 - 5) = 0$$ $$-4 + \lambda(-5) = 0$$ $$-5\lambda = 4 \implies \lambda = -\frac{4}{5}$$

Step 4: Substitute $\lambda$ back into the family equation.
Replace $\lambda$ with $-4/5$ in the original equation from
Step 1: $$(x + 2y - z - 3) - \frac{4}{5}(x + y - 3z - 5) = 0$$ Multiply the entire equation by 5 to clear the fraction: $$5(x + 2y - z - 3) - 4(x + y - 3z - 5) = 0$$

Step 5: Expand and simplify into standard form.
Distribute the scalars across the parentheses: $$(5x + 10y - 5z - 15) - (4x + 4y - 12z - 20) = 0$$ Combine the like terms for $x$, $y$, $z$, and constants: $$x + 6y + 7z + 5 = 0$$ Hence the correct answer is (E) $x+6y+7z+5=0$.