A unit vector a→ is parallel to the line of intersection of the plane determined by the vectorsi^,i^+j^ and plane determined by the vectors i^-j^and i^+k^ . The cosine of the angle between a→ and i^-2j^+2k^ is equal to

n1=ijk100110=i(0)-j(0)+k(1)=k n2=ijk1-10101=i(-1)-j(1)+k(1) n1×n2=ijk001-1-11=i(1)-j(1)+k(0) =i-j a=λn1×n2=λ(i-j) b=i-2j+2k cos θ=1+22 1+4+4=12 ⇒ θ=π4

A unit vector a→ is parallel to the line of intersection of the plane determined by the vectorsi^,i^+j^ and plane determined by the vectors i^-j^and i^+k^ . The cosine of the angle between a→ and i^-2j^+2k^ is equal to

n1=ijk100110=i(0)-j(0)+k(1)=k n2=ijk1-10101=i(-1)-j(1)+k(1) n1×n2=ijk001-1-11=i(1)-j(1)+k(0) =i-j a=λn1×n2=λ(i-j) b=i-2j+2k cos θ=1+22 1+4+4=12 ⇒ θ=π4

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A unit vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i} + \hat{j}$ and plane determined by the vectors $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$ . The cosine of the angle between $\vec{a}$ and $\hat{i} - 2 \hat{j} + 2 \hat{k}$ is equal to

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$\frac{1}{\sqrt{2}}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{\sqrt{3}}{2}$

answer is A.

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Detailed Solution

$n_{1} = |\begin{matrix} i & j & k \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{matrix}| = i (0) - j (0) + k (1) = k n_{2} = |\begin{matrix} i & j & k \\ 1 & - 1 & 0 \\ 1 & 0 & 1 \end{matrix}| = i (- 1) - j (1) + k (1) n_{1} \times n_{2} = |\begin{matrix} i & j & k \\ 0 & 0 & 1 \\ - 1 & - 1 & 1 \end{matrix}| = i (1) - j (1) + k (0) = i - j a = λ (n_{1} \times n_{2}) = λ (i - j) b = i - 2 j + 2 k \cos θ = \frac{1 + 2}{\sqrt{2} \sqrt{1 + 4 + 4}} = \frac{1}{\sqrt{2}} \Rightarrow θ = \frac{π}{4}$