Question

Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}, \hat{i}+\hat{k}$ and $\pi_2$ be the plane determined by the vectors $\hat{j}-\hat{k}, \hat{k}-\hat{i}$. Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\vec{b} = \hat{i}+\hat{j}-\hat{k}$ then the angle between the vectors $\vec{a}$ and $\vec{b}$ is

• A. $\text{Cos}^{-1}\left(\frac{2}{\sqrt{3}}\right)$
• B. $\frac{\pi}{2}$
• C. $\text{Cos}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• D. $\text{Cos}^{-1}\left(\frac{\sqrt{2}}{3}\right)$

Hint:

A vector parallel to the line of intersection of two planes can be found in two ways: (1) as the cross product of the normal vectors of the planes, or (2) by finding a common vector in the span of the direction vectors of each plane. If your result from a standard method does not match any options, double-check your calculations. If it's still inconsistent, consider a potential typo in the problem statement, especially a simple sign error.

Answer 1

The Correct Option is B

Step 1: Find Normal Vectors to the Planes:

The normal to π1 is n1 = (i + j) × (i + k).

n₁ = | i   j   k |
     | 1   1   0 |
     | 1   0   1 |

n1 = i(1 - 0) - j(1 - 0) + k(0 - 1) = i - j - k

The normal to π2 is n2 = (j - k) × (k - i).

n₂ = | i   j   k |
     | 0   1  -1 |
     | -1  0   1 |

n2 = i(1 - 0) - j(0 - 1) + k(0 + 1) = i + j + k

Step 2: Find Vector Parallel to Intersection Line:

The line of intersection is perpendicular to both normals.

a = n₁ × n₂ = | i   j   k |
              | 1  -1  -1 |
              | 1   1   1 |

a = i(-1 + 1) - j(1 + 1) + k(1 + 1) = 0i - 2j + 2k

We can simplify the direction vector to a = -j + k.

Step 3: Calculate Angle with b:

b = i + j - k

Angle θ is given by:

cos θ = |a · b| / (|a| |b|)

a · b = (0)(1) + (-1)(1) + (1)(-1) = -2

|a| = √(0 + 1 + 1) = √2

|b| = √(1 + 1 + 1) = √3

cos θ = |-2| / (√2 × √3) = 2 / √6 = 2√6 / 6 = √(2 / 3)

θ = cos-1(√(2 / 3))