Question

The angle between the planes $x+y+z=1$ and $x-2y+3z=1$ is

• A. $\cos^{-1}(\frac{2}{\sqrt{42}})$
• B. $\cos^{-1}(\frac{5}{\sqrt{42}})$
• C. $\cos^{-1}(\frac{3}{\sqrt{42}})$
• D. $\cos^{-1}(\frac{1}{\sqrt{42}})$
• E. $\cos^{-1}(\frac{4}{\sqrt{42}})$

Hint:

Vector Tip: The absolute value is used in the numerator because the angle between two intersecting planes is conventionally given as the acute angle ($0 \le \theta \le 90^\circ$).

Answer 1

The Correct Option is A

Concept:
The angle $\theta$ between two planes is equivalent to the angle between their respective normal vectors. The normal vector of a plane $Ax + By + Cz = D$ is simply $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. We use the dot product formula $\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}$ to find the angle.

Step 1: Extract the normal vectors.
For the first plane $x + y + z = 1$, the normal vector is: $$\vec{n_1} = (1, 1, 1)$$ For the second plane $x - 2y + 3z = 1$, the normal vector is: $$\vec{n_2} = (1, -2, 3)$$

Step 2: Calculate the dot product of the normal vectors.
Multiply the corresponding components and add them: $$\vec{n_1} \cdot \vec{n_2} = (1)(1) + (1)(-2) + (1)(3)$$ $$\vec{n_1} \cdot \vec{n_2} = 1 - 2 + 3 = 2$$

Step 3: Calculate the magnitudes of the normal vectors.
Find the length of both vectors using the Pythagorean distance formula: $$|\vec{n_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$ $$|\vec{n_2}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$

Step 4: Substitute into the cosine formula.
Plug the dot product and magnitudes into the geometric angle formula: $$\cos\theta = \frac{|2|}{\sqrt{3} \times \sqrt{14}}$$ $$\cos\theta = \frac{2}{\sqrt{42}}$$

Step 5: Solve for the angle $\theta$.
Take the inverse cosine of both sides to isolate $\theta$: $$\theta = \cos^{-1}\left(\frac{2}{\sqrt{42}}\right)$$ Hence the correct answer is (A) $\cos^{-1(\frac{2}{\sqrt{42}})$}.