Related Formula
Cross product for vector perpendicular direction alignment:
u = b × c$$\vec{u} = \vec{b} \times \vec{c}$$
Angle projection formula:
θ = a · v| a|| v|$$\cos\theta = \frac{\vec{a} \cdot \vec{v}}{|\vec{a}||\vec{v}|}$$
Core Logic
Compute cross product of b$\vec{b}$ and c$\vec{c}$:
b × c = vmatrix i & j & k 1 & -2 & 3 2 & 3 & -1 vmatrix = -7 i + 7 j + 7 k = -7( i - j - k)$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = -7\hat{i} + 7\hat{j} + 7\hat{k} = -7(\hat{i} - \hat{j} - \hat{k})$$
Hence, unit vector a$\hat{a}$ matches form:
a = ± i - j - k√(3)$$\hat{a} = \pm \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$$
Step 1: Isolate Core Angle Direction
Check conditions against vector v = i + j + k$\vec{v} = \hat{i} + \hat{j} + \hat{k}$:
Using a = i - j - k√(3)$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$:
θ = 1 - 1 - 1√(3)√(3) = -(1)/(3)$$\cos\theta = \frac{1 - 1 - 1}{\sqrt{3}\sqrt{3}} = -\frac{1}{3}$$
This confirms the direction for a$\hat{a}$.
Step 2: Solve for Unknown Scalar Variable
Now compute angle with vector i + α j + k$\hat{i} + \alpha\hat{j} + \hat{k}$ for θ = (π)/(3)$\theta = \frac{\pi}{3}$:
(π)/(3) = 1√(3) · 1 - α - 1√(2 + α²)$$\cos\frac{\pi}{3} = \frac{1}{\sqrt{3}} \cdot \frac{1 - \alpha - 1}{\sqrt{2 + \alpha^2}}$$
(1)/(2) = -α√(3)√(α² + 2)$$\frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$$
Since left hand side is positive, α$\alpha$ must be strictly negative. Squaring both sides:
(1)/(4) = (α²)/(3(α² + 2)) 3α² + 6 = 4α² α² = 6$$\frac{1}{4} = \frac{\alpha^2}{3(\alpha^2 + 2)} \implies 3\alpha^2 + 6 = 4\alpha^2 \implies \alpha^2 = 6$$
Since α < 0$\alpha < 0$, α = -√(6)$\alpha = -\sqrt{6}$.
Pattern Recognition
Keep strict track of signs when dealing with algebra containing square roots. Checking value constraints early on allows you to drop phantom positive/negative branches seamlessly.
Chapter Mix
Class 12 Mathematics: Vector Algebra
