Let hatmathbfa$\hat{\mathbf{a}}$ be a unit vector perpendicular to the vectors vecmathsfb = hatmathsfi -2hatmathsfj +3hatmathsfk$\vec{\mathsf{b}} = \hat{\mathsf{i}} -2\hat{\mathsf{j}} +3\hat{\mathsf{k}}$ and vecmathbfc = 2hatmathbfi +3hatmathbfj -hatmathbfk$\vec{\mathbf{c}} = 2\hat{\mathbf{i}} +3\hat{\mathbf{j}} -\hat{\mathbf{k}}$, and makes an angle of cos^-1left(-frac13right)$\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector hatmathrmi +hatmathrmj +hatmathrmk$\hat{\mathrm{i}} +\hat{\mathrm{j}} +\hat{\mathrm{k}}$. If hatmathbfa$\hat{\mathbf{a}}$ makes an angle of fracpi3$\frac{\pi}{3}$ with the vector hatmathrmi +alpha hatmathrmj +hatmathrmk$\hat{\mathrm{i}} +\alpha \hat{\mathrm{j}} +\hat{\mathrm{k}}$, then the value of alpha$\alpha$ is :

### Related Formula Cross product for vector perpendicular direction alignment: vecu = vecb times vecc$$\vec{u} = \vec{b} \times \vec{c}$$ Angle projection formula: costheta = fracveca cdot vecv|veca||vecv|$$\cos\theta = \frac{\vec{a} \cdot \vec{v}}{|\vec{a}||\vec{v}|}$$ ### Core Logic Compute cross product of vecb$\vec{b}$ and vecc$\vec{c}$: vecb times vecc = beginvmatrix hati & hatj & hatk \\ 1 & -2 & 3 \\ 2 & 3 & -1 endvmatrix = -7hati + 7hatj + 7hatk = -7(hati - hatj - hatk)$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = -7\hat{i} + 7\hat{j} + 7\hat{k} = -7(\hat{i} - \hat{j} - \hat{k})$$ Hence, unit vector hata$\hat{a}$ matches form: hata = pm frachati - hatj - hatksqrt3$$\hat{a} = \pm \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$$ ### Step 1: Isolate Core Angle Direction Check conditions against vector vecv = hati + hatj + hatk$\vec{v} = \hat{i} + \hat{j} + \hat{k}$: Using hata = frachati - hatj - hatksqrt3$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$: costheta = frac1 - 1 - 1sqrt3sqrt3 = -frac13$$\cos\theta = \frac{1 - 1 - 1}{\sqrt{3}\sqrt{3}} = -\frac{1}{3}$$ This confirms the direction for hata$\hat{a}$. ### Step 2: Solve for Unknown Scalar Variable Now compute angle with vector hati + alphahatj + hatk$\hat{i} + \alpha\hat{j} + \hat{k}$ for theta = fracpi3$\theta = \frac{\pi}{3}$: cosfracpi3 = frac1sqrt3 cdot frac1 - alpha - 1sqrt2 + alpha^2$$\cos\frac{\pi}{3} = \frac{1}{\sqrt{3}} \cdot \frac{1 - \alpha - 1}{\sqrt{2 + \alpha^2}}$$ frac12 = frac-alphasqrt3sqrtalpha^2 + 2$$\frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$$ Since left hand side is positive, alpha$\alpha$ must be strictly negative. Squaring both sides: frac14 = fracalpha^23(alpha^2 + 2) implies 3alpha^2 + 6 = 4alpha^2 implies alpha^2 = 6$$\frac{1}{4} = \frac{\alpha^2}{3(\alpha^2 + 2)} \implies 3\alpha^2 + 6 = 4\alpha^2 \implies \alpha^2 = 6$$ Since alpha < 0$\alpha < 0$, alpha = -sqrt6$\alpha = -\sqrt{6}$. ### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra