Question

If $\vec{a}$, $\vec{b}$, $\vec{c}$ are three vectors with magnitudes $\sqrt{3}$, $1$, $2$ respectively, such that $\vec{a} \times (\vec{a} \times \vec{c}) + 3\vec{b} = \vec{0}$, if $\theta$ is the angle between $\vec{a}$ and $\vec{c}$, then $\sec^2 \theta$ is

• A. 1
• B. $\frac{3}{2}$
• C. $\frac{4}{3}$
• D. $\frac{2}{\sqrt{3}}$

Hint:

Vector Tip:The "BAC-CAB" rule for the vector triple product: $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B})$.

Answer 1

The Correct Option is C

Concept: Vector Algebra - Vector Triple Product and Dot Product.

Step 1: Apply the Vector Triple Product expansion. The given equation is $\overline{a}\times(\overline{a}\times\overline{c})+3\overline{b}=\overline{0}$. We use the standard vector triple product expansion identity $\overline{A}\times(\overline{B}\times\overline{C}) = (\overline{A}\cdot\overline{C})\overline{B} - (\overline{A}\cdot\overline{B})\overline{C}$. Applying this to our equation gives: $(\overline{a}\cdot\overline{c})\overline{a} - (\overline{a}\cdot\overline{a})\overline{c} + 3\overline{b} = \overline{0}$.

Step 2: Simplify the equation using given magnitudes. We are given the magnitudes $|\overline{a}|=\sqrt{3}$, $|\overline{b}|=1$, and $|\overline{c}|=2$. We know that $\overline{a}\cdot\overline{a} = |\overline{a}|^2 = (\sqrt{3})^2 = 3$. Substituting this into our expanded equation yields: $(\overline{a}\cdot\overline{c})\overline{a} - 3\overline{c} + 3\overline{b} = \overline{0}$.

Step 3: Isolate vector $\overline{b}$ and square both sides. To utilize the magnitude of $\overline{b}$, let's isolate it: $(\overline{a}\cdot\overline{c})\overline{a} - 3\overline{c} = -3\overline{b}$. Now, take the square of the magnitude of both sides: $|(\overline{a}\cdot\overline{c})\overline{a} - 3\overline{c}|^2 = |-3\overline{b}|^2$.

Step 4: Expand the squared expression and solve for the dot product. Expanding the left side gives: $(\overline{a}\cdot\overline{c})^2|\overline{a}|^2 + 9|\overline{c}|^2 - 2(3)(\overline{a}\cdot\overline{c})(\overline{a}\cdot\overline{c}) = 9|\overline{b}|^2$. This simplifies to: $(\overline{a}\cdot\overline{c})^2|\overline{a}|^2 + 9|\overline{c}|^2 - 6(\overline{a}\cdot\overline{c})^2 = 9|\overline{b}|^2$. Now, substitute the known squared magnitudes ($|\overline{a}|^2=3, |\overline{c}|^2=4, |\overline{b}|^2=1$): $3(\overline{a}\cdot\overline{c})^2 + 9(4) - 6(\overline{a}\cdot\overline{c})^2 = 9(1)$. This simplifies to $36 - 3(\overline{a}\cdot\overline{c})^2 = 9$, which means $3(\overline{a}\cdot\overline{c})^2 = 27$, and therefore $(\overline{a}\cdot\overline{c})^2 = 9$. This gives $\overline{a}\cdot\overline{c} = \pm 3$.

Step 5: Calculate $\sec^{2}\theta$ using the dot product formula. By definition, $\overline{a}\cdot\overline{c} = |\overline{a}||\overline{c}|\cos\theta$. Substituting the knowns: $\sqrt{3}(2)\cos\theta = \pm 3$. Solving for $\cos\theta$ gives $\cos\theta = \pm \frac{3}{2\sqrt{3}} = \pm \frac{\sqrt{3}}{2}$. Since $\sec\theta$ is the reciprocal of $\cos\theta$, we have $\sec\theta = \pm \frac{2}{\sqrt{3}}$. Squaring this value gives $\sec^2\theta = (\pm \frac{2}{\sqrt{3}})^2 = \frac{4}{3}$. $$ \therefore \text{The value of } \sec^2\theta \text{ is } \frac{4}{3}. $$