Question

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=4$ and $\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}=0$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$, then $\vec{a}$ is equal to

• A. $\pm\frac{1}{\sqrt{3}}(\vec{b}\times\vec{c})$
• B. $\pm\frac{1}{2\sqrt{3}}(\vec{b}\times\vec{c})$
• C. $\pm\frac{1}{3\sqrt{2}}(\vec{b}\times\vec{c})$
• D. $\pm\frac{1}{3\sqrt{3}}(\vec{b}\times\vec{c})$
• E. $\pm\frac{1}{2\sqrt{2}}(\vec{b}\times\vec{c})$

Hint:

Math Tip: When a vector $\vec{x}$ is mutually orthogonal to two vectors $\vec{y}$ and $\vec{z}$, the fastest way to represent it is $\vec{x} = \pm \frac{|\vec{x}|}{|\vec{y} \times \vec{z}|} (\vec{y} \times \vec{z})$.

Answer 1

The Correct Option is D

Concept:
Vectors - Cross Product and Orthogonality.
If a vector is perpendicular to two other non-collinear vectors, it must be parallel to their cross product.
Step 1: Establish the relationship between vectors.
We are given $\vec{a}\cdot\vec{b}=0$ and $\vec{a}\cdot\vec{c}=0$.
This implies that vector $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$.
Therefore, $\vec{a}$ must be parallel to the normal of the plane containing $\vec{b}$ and $\vec{c}$, which is given by their cross product $(\vec{b}\times\vec{c})$.
We can write: $\vec{a} = \lambda (\vec{b}\times\vec{c})$, where $\lambda$ is a scalar constant.
Step 2: Take the magnitude of both sides.
$$ |\vec{a}| = |\lambda| |\vec{b}\times\vec{c}| $$
Step 3: Calculate the magnitude of the cross product $|\vec{b}\times\vec{c}|$.
The formula for the magnitude of a cross product is $|\vec{b}\times\vec{c}| = |\vec{b}||\vec{c}|\sin\theta$.
Substitute the given values ($|\vec{b}|=3, |\vec{c}|=4$, and $\theta = \frac{\pi}{3}$): $$ |\vec{b}\times\vec{c}| = (3)(4)\sin\left(\frac{\pi}{3}\right) $$ $$ |\vec{b}\times\vec{c}| = 12 \left(\frac{\sqrt{3}}{2}\right) = 6\sqrt{3} $$
Step 4: Solve for the scalar $\lambda$.
Substitute $|\vec{a}| = 2$ and $|\vec{b}\times\vec{c}| = 6\sqrt{3}$ back into the magnitude equation: $$ 2 = |\lambda| (6\sqrt{3}) $$ $$ |\lambda| = \frac{2}{6\sqrt{3}} = \frac{1}{3\sqrt{3}} $$ This means $\lambda$ can be positive or negative: $$ \lambda = \pm\frac{1}{3\sqrt{3}} $$
Step 5: Construct the final expression for $\vec{a}$.
Substitute $\lambda$ back into our original assumption $\vec{a} = \lambda (\vec{b}\times\vec{c})$: $$ \vec{a} = \pm\frac{1}{3\sqrt{3}}(\vec{b}\times\vec{c}) $$