We are given the vectors
a ⃗ = i ^ + λ j ^ − 3 k ^ and b ⃗ = 3 i ^ − j ^ + 2 k ^ , \vec{a} = \hat{i} + \lambda\,\hat{j} - 3\,\hat{k} \quad \text{and} \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}, a = i ^ + λ j ^ − 3 k ^ and b = 3 i ^ − j ^ + 2 k ^ ,
with \lambda > 0 , and we are told that the vectors a ⃗ + b ⃗ \vec{a}+\vec{b} a + b and a ⃗ − b ⃗ \vec{a}-\vec{b} a − b are perpendicular. Our goal is to find ( 14 cos θ ) 2 (14 \cos\theta)^2 ( 14 cos θ ) 2 , where θ \theta θ is the angle between a ⃗ \vec{a} a and b ⃗ \vec{b} b .
Step 1. Use the Perpendicularity Condition Two vectors are perpendicular if their dot product is zero. So, we require:
( a ⃗ + b ⃗ ) ⋅ ( a ⃗ − b ⃗ ) = 0. (\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0. ( a + b ) ⋅ ( a − b ) = 0.
First, compute a ⃗ + b ⃗ \vec{a}+\vec{b} a + b and a ⃗ − b ⃗ \vec{a}-\vec{b} a − b :
a ⃗ + b ⃗ = ( i ^ + 3 i ^ ) + ( λ j ^ − j ^ ) + ( − 3 k ^ + 2 k ^ ) = 4 i ^ + ( λ − 1 ) j ^ − k ^ . \vec{a}+\vec{b} = (\hat{i}+3\,\hat{i})+(\lambda\,\hat{j}-\hat{j})+(-3\,\hat{k}+2\,\hat{k}) = 4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}. a + b = ( i ^ + 3 i ^ ) + ( λ j ^ − j ^ ) + ( − 3 k ^ + 2 k ^ ) = 4 i ^ + ( λ − 1 ) j ^ − k ^ .
a ⃗ − b ⃗ = ( i ^ − 3 i ^ ) + ( λ j ^ + j ^ ) + ( − 3 k ^ − 2 k ^ ) = − 2 i ^ + ( λ + 1 ) j ^ − 5 k ^ . \vec{a}-\vec{b} = (\hat{i}-3\,\hat{i})+(\lambda\,\hat{j}+ \hat{j})+(-3\,\hat{k}-2\,\hat{k}) = -2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}. a − b = ( i ^ − 3 i ^ ) + ( λ j ^ + j ^ ) + ( − 3 k ^ − 2 k ^ ) = − 2 i ^ + ( λ + 1 ) j ^ − 5 k ^ .
Now, take their dot product:
( 4 i ^ + ( λ − 1 ) j ^ − k ^ ) ⋅ ( − 2 i ^ + ( λ + 1 ) j ^ − 5 k ^ ) (4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}) \cdot (-2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}) ( 4 i ^ + ( λ − 1 ) j ^ − k ^ ) ⋅ ( − 2 i ^ + ( λ + 1 ) j ^ − 5 k ^ )
= 4 ( − 2 ) + ( λ − 1 ) ( λ + 1 ) + ( − 1 ) ( − 5 ) = − 8 + ( λ 2 − 1 ) + 5. = 4(-2) + (\lambda-1)(\lambda+1) + (-1)(-5) = -8 + (\lambda^2 - 1) + 5. = 4 ( − 2 ) + ( λ − 1 ) ( λ + 1 ) + ( − 1 ) ( − 5 ) = − 8 + ( λ 2 − 1 ) + 5.
= λ 2 − 4. = \lambda^2 - 4. = λ 2 − 4.
Setting the dot product to zero:
λ 2 − 4 = 0 ⟹ λ 2 = 4. \lambda^2 - 4 = 0 \quad \Longrightarrow \quad \lambda^2 = 4. λ 2 − 4 = 0 ⟹ λ 2 = 4.
Since \lambda > 0 , we have:
λ = 2. \lambda = 2. λ = 2.
Step 2. Find the Angle Between a ⃗ \vec{a} a and b ⃗ \vec{b} b Now that we know λ = 2 \lambda = 2 λ = 2 , the vectors become:
a ⃗ = i ^ + 2 j ^ − 3 k ^ , b ⃗ = 3 i ^ − j ^ + 2 k ^ . \vec{a} = \hat{i} + 2\,\hat{j} - 3\,\hat{k}, \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}. a = i ^ + 2 j ^ − 3 k ^ , b = 3 i ^ − j ^ + 2 k ^ .
The cosine of the angle θ \theta θ between a ⃗ \vec{a} a and b ⃗ \vec{b} b is given by:
cos θ = a ⃗ ⋅ b ⃗ ∥ a ⃗ ∥ ∥ b ⃗ ∥ . \cos\theta = \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\| \|\vec{b}\|}. cos θ = ∥ a ∥∥ b ∥ a ⋅ b .
Calculate a ⃗ ⋅ b ⃗ \vec{a} \cdot \vec{b} a ⋅ b :
a ⃗ ⋅ b ⃗ = ( 1 ) ( 3 ) + ( 2 ) ( − 1 ) + ( − 3 ) ( 2 ) = 3 − 2 − 6 = − 5. \vec{a}\cdot\vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5. a ⋅ b = ( 1 ) ( 3 ) + ( 2 ) ( − 1 ) + ( − 3 ) ( 2 ) = 3 − 2 − 6 = − 5.
Calculate the magnitudes:
∥ a ⃗ ∥ = 1 2 + 2 2 + ( − 3 ) 2 = 1 + 4 + 9 = 14 . \|\vec{a}\| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}. ∥ a ∥ = 1 2 + 2 2 + ( − 3 ) 2 = 1 + 4 + 9 = 14 .
∥ b ⃗ ∥ = 3 2 + ( − 1 ) 2 + 2 2 = 9 + 1 + 4 = 14 . \|\vec{b}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}. ∥ b ∥ = 3 2 + ( − 1 ) 2 + 2 2 = 9 + 1 + 4 = 14 .
Thus,
cos θ = − 5 14 ⋅ 14 = − 5 14 . \cos\theta = \frac{-5}{\sqrt{14}\cdot\sqrt{14}} = \frac{-5}{14}. cos θ = 14 ⋅ 14 − 5 = 14 − 5 .
Step 3. Compute ( 14 cos θ ) 2 (14\cos\theta)^2 ( 14 cos θ ) 2 ( 14 cos θ ) 2 = ( 14 ⋅ − 5 14 ) 2 = ( − 5 ) 2 = 25. (14\cos\theta)^2 = \left(14\cdot\frac{-5}{14}\right)^2 = (-5)^2 = 25. ( 14 cos θ ) 2 = ( 14 ⋅ 14 − 5 ) 2 = ( − 5 ) 2 = 25.
Final Answer The value of ( 14 cos θ ) 2 (14\cos\theta)^2 ( 14 cos θ ) 2 is:
25 . \boxed{25}. 25 .