Question

If $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors and $\theta$ is the angle between $\vec{a}$ and $\vec{c}$ and $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$, then $|\vec{a} \times \vec{c}|$ is equal to

• A. $\frac{\sqrt{15}}{2}$
• B. $\frac{\sqrt{15}}{4}$
• C. $\sqrt{15}$
• D. $\frac{\sqrt{15}}{3}$

Hint:

Vector Tip: Whenever a relation like $\vec{x} + \vec{y} + \vec{z} = 0$ is given and you need an angle or product involving two of them, isolate the third vector and square both sides.

Answer 1

The Correct Option is B

Concept: Vector Algebra - Magnitudes, Dot Product, and Cross Product of Vectors.

Step 1: Isolate the term not involved in the cross product. We are given the vector equation $\overline{a}+2\overline{b}+2\overline{c}=\overline{0}$. Since we need to find the value of $|\overline{a}\times\overline{c}|$, which involves vectors $\overline{a}$ and $\overline{c}$, let's isolate vector $\overline{b}$ on one side: $\overline{a}+2\overline{c}=-2\overline{b}$.

Step 2: Square both sides to utilize the dot product. Squaring both sides of the equation gives $|\overline{a}+2\overline{c}|^2 = |-2\overline{b}|^2$. Expanding the left side using the property $|\overline{x}+\overline{y}|^2 = |\overline{x}|^2 + 2(\overline{x}\cdot\overline{y}) + |\overline{y}|^2$ gives: $|\overline{a}|^2 + 4(\overline{a}\cdot\overline{c}) + 4|\overline{c}|^2 = 4|\overline{b}|^2$.

Step 3: Substitute unit vector magnitudes and solve for the dot product. Since $\overline{a}$, $\overline{b}$, and $\overline{c}$ are unit vectors, their magnitudes are 1 ($|\overline{a}|=1, |\overline{b}|=1, |\overline{c}|=1$). Substituting these values: $(1)^2 + 4(\overline{a}\cdot\overline{c}) + 4(1)^2 = 4(1)^2$. This simplifies to $1 + 4(\overline{a}\cdot\overline{c}) + 4 = 4$, which means $4(\overline{a}\cdot\overline{c}) = -1$. Using the dot product definition $\overline{a}\cdot\overline{c} = |\overline{a}||\overline{c}|\cos\theta$, we get $4(1)(1)\cos\theta = -1$, leading to $\cos\theta = -\frac{1}{4}$.

Step 4: Calculate $\sin\theta$ from $\cos\theta$. We know the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$. Therefore, $\sin\theta = \sqrt{1 - \cos^2\theta}$. Substituting $\cos\theta = -\frac{1}{4}$ yields $\sin\theta = \sqrt{1 - (-\frac{1}{4})^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$. (We take the positive root because the angle between two vectors is defined in $[0, \pi]$, where sine is always positive).

Step 5: Calculate the magnitude of the cross product. The magnitude of the cross product is given by the formula $|\overline{a}\times\overline{c}| = |\overline{a}||\overline{c}|\sin\theta$. Substituting the known values: $|\overline{a}\times\overline{c}| = (1)(1)(\frac{\sqrt{15}}{4}) = \frac{\sqrt{15}}{4}$. $$ \therefore \text{The value of } |\overline{a}\times\overline{c}| \text{ is } \frac{\sqrt{15}}{4}. $$