Question

$\overline{b}$ and $\overline{c}$ are non-collinear vectors and $\overline{a}$ is a vector such that $(\overline{c}\cdot\overline{c})\overline{a}=\overline{c}$. If $\overline{a}\times(\overline{b}\times\overline{c})+(\overline{a}\cdot\overline{b})\overline{b}=(4-2\beta-sin~\alpha)\overline{b}+(\beta^{2}-1)\overline{c}$, then the values of the scalars $\alpha$ and $\beta$ are

• A. \(\beta=2, \alpha=n\pi+\frac{\pi}{2}, n\in z \)
• B. \(\beta=-1, \alpha=2n\pi+\frac{\pi}{4}, n\in z \)
• C. \(\beta=1, \alpha=(2n+1)\frac{\pi}{2}, n\in z \)
• D. \(\beta=1, \alpha=2n\pi+\frac{\pi}{2}, n\in z \)

Hint:

When a vector equation equates linear combinations of non-collinear vectors, the coefficients must be identical. Also, remember the range of trigonometric functions to constrain variables.

Answer 1

The Correct Option is D

Concept: The problem requires an understanding of the vector triple product identity and the linear independence of vectors. We use the identity: \[ \overline{a}\times(\overline{b}\times\overline{c}) = (\overline{a}\cdot\overline{c})\overline{b} - (\overline{a}\cdot\overline{b})\overline{c} \]

Step 1: Expand the vector triple product in the given equation.
Given: $\overline{a}\times(\overline{b}\times\overline{c})+(\overline{a}\cdot\overline{b})\overline{b}=(4-2\beta-sin~\alpha)\overline{b}+(\beta^{2}-1)\overline{c}$
Applying the identity to $\overline{a}\times(\overline{b}\times\overline{c})$:
$[(\overline{a}\cdot\overline{c})\overline{b} - (\overline{a}\cdot\overline{b})\overline{c}] + (\overline{a}\cdot\overline{b})\overline{b} = (4-2\beta-sin~\alpha)\overline{b}+(\beta^{2}-1)\overline{c}$

Step 2: Group the vectors \(\overline{b}\) and \(\overline{c}\) on the left side.
Collect terms involving $\overline{b}$ and terms involving $\overline{c}$:
$(\overline{a}\cdot\overline{c} + \overline{a}\cdot\overline{b})\overline{b} - (\overline{a}\cdot\overline{b})\overline{c} = (4-2\beta-sin~\alpha)\overline{b}+(\beta^{2}-1)\overline{c}$

Step 3: Evaluate dot products using the condition \( (\overline{c}\cdot\overline{c})\overline{a}=\overline{c} \).
Since $(\overline{c}\cdot\overline{c})\overline{a}=\overline{c}$, it follows that $\overline{a} = \frac{\overline{c}}{|\overline{c}|^2}$.
Substituting this into the dot products:
$\overline{a}\cdot\overline{c} = \left(\frac{\overline{c}}{|\overline{c}|^2}\right)\cdot\overline{c} = \frac{\overline{c}\cdot\overline{c}}{|\overline{c}|^2} = 1$
$\overline{a}\cdot\overline{b} = \left(\frac{\overline{c}}{|\overline{c}|^2}\right)\cdot\overline{b} = \frac{\overline{c}\cdot\overline{b}}{|\overline{c}|^2}$

Step 4: Equate the coefficients of the non-collinear vectors \(\overline{b}\) and \(\overline{c}\).
Comparing coefficients:
For $\overline{c}$: $-\frac{\overline{c}\cdot\overline{b}}{|\overline{c}|^2} = \beta^2 - 1$
For $\overline{b}$: $1 + \frac{\overline{c}\cdot\overline{b}}{|\overline{c}|^2} = 4 - 2\beta - \sin\alpha$

Step 5: Solve the resulting system of equations for \(\beta\) and \(\alpha\).
Adding the two coefficient equations yields:
$1 = (4 - 2\beta - \sin\alpha) + (\beta^2 - 1)$
$1 = \beta^2 - 2\beta + 3 - \sin\alpha$
$\sin\alpha = \beta^2 - 2\beta + 2 = (\beta-1)^2 + 1$
Since the maximum value of $\sin\alpha$ is 1, this requires $(\beta-1)^2 = 0$, implying $\beta = 1$.
Substituting $\beta = 1$ back gives $\sin\alpha = 1$, which solves to $\alpha = 2n\pi + \frac{\pi}{2}$. center minipage0.4

$\beta = 1, \alpha = 2n\pi + \frac{\pi}{2}, n \in z$ minipage center