Question

Let $\bar{a}$ and $\bar{b}$ be two unit vectors such that $\bar{a} \cdot \bar{b}=0$. For some $x, y \in R$, let $\bar{c} = x\bar{a} + y\bar{b} + (\bar{a} \times \bar{b})$. If $|\bar{c}|=2$ and the vector $\bar{c}$ is inclined at the same angle $\alpha$ to both $\bar{a}$ and $\bar{b}$, then the value of $x^2+y^2+8\cos^2\alpha$ is equal to:

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (D)

6

Answer 2

Given that $\bar{a}$ and $\bar{b}$ are unit vectors with $\bar{a} \cdot \bar{b} = 0$, they form an orthonormal set. Let $\bar{d} = \bar{a} \times \bar{b}$. Since $\bar{a}$ and $\bar{b}$ are orthogonal unit vectors, $|\bar{d}| = |\bar{a}||\bar{b}|\sin(\pi/2) = 1 \cdot 1 \cdot 1 = 1$. Also, $\bar{d}$ is orthogonal to both $\bar{a}$ and $\bar{b}$, meaning $\bar{d} \cdot \bar{a} = 0$ and $\bar{d} \cdot \bar{b} = 0$. Thus, $\{\bar{a}, \bar{b}, \bar{d}\}$ forms an orthonormal basis.

We are given $\bar{c} = x\bar{a} + y\bar{b} + \bar{d}$ and $|\bar{c}|=2$. Squaring the magnitude of $\bar{c}$: $|\bar{c}|^2 = (x\bar{a} + y\bar{b} + \bar{d}) \cdot (x\bar{a} + y\bar{b} + \bar{d})$ Using the orthonormality of $\bar{a}, \bar{b}, \bar{d}$: $|\bar{c}|^2 = x^2|\bar{a}|^2 + y^2|\bar{b}|^2 + |\bar{d}|^2 + 2xy(\bar{a} \cdot \bar{b}) + 2x(\bar{a} \cdot \bar{d}) + 2y(\bar{b} \cdot \bar{d})$ $|\bar{c}|^2 = x^2(1) + y^2(1) + 1^2 + 2xy(0) + 2x(0) + 2y(0)$ $|\bar{c}|^2 = x^2 + y^2 + 1$ Since $|\bar{c}| = 2$, we have $|\bar{c}|^2 = 4$. So, $x^2 + y^2 + 1 = 4$, which implies $x^2 + y^2 = 3$.

Now, let $\alpha$ be the angle between $\bar{c}$ and $\bar{a}$. The cosine of this angle is given by: $\cos\alpha = \frac{\bar{c} \cdot \bar{a}}{|\bar{c}||\bar{a}|}$ $\cos\alpha = \frac{(x\bar{a} + y\bar{b} + \bar{d}) \cdot \bar{a}}{2 \cdot 1}$ $\cos\alpha = \frac{x|\bar{a}|^2 + y(\bar{b} \cdot \bar{a}) + (\bar{d} \cdot \bar{a})}{2}$ $\cos\alpha = \frac{x(1) + y(0) + 0}{2} = \frac{x}{2}$ Thus, $x = 2\cos\alpha$.

Similarly, let $\alpha$ be the angle between $\bar{c}$ and $\bar{b}$. $\cos\alpha = \frac{\bar{c} \cdot \bar{b}}{|\bar{c}||\bar{b}|}$ $\cos\alpha = \frac{(x\bar{a} + y\bar{b} + \bar{d}) \cdot \bar{b}}{2 \cdot 1}$ $\cos\alpha = \frac{x(\bar{a} \cdot \bar{b}) + y|\bar{b}|^2 + (\bar{d} \cdot \bar{b})}{2}$ $\cos\alpha = \frac{x(0) + y(1) + 0}{2} = \frac{y}{2}$ Thus, $y = 2\cos\alpha$.

Now we substitute $x = 2\cos\alpha$ and $y = 2\cos\alpha$ into the equation $x^2 + y^2 = 3$: $(2\cos\alpha)^2 + (2\cos\alpha)^2 = 3$ $4\cos^2\alpha + 4\cos^2\alpha = 3$ $8\cos^2\alpha = 3$

We need to find the value of $x^2+y^2+8\cos^2\alpha$. We found $x^2+y^2 = 3$ and $8\cos^2\alpha = 3$. Therefore, $x^2+y^2+8\cos^2\alpha = 3 + 3 = 6$.