Question

If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \( |\vec{a}+\vec{b}+\vec{c}|=1 \), \( \vec{c}=\lambda(\vec{a}\times\vec{b}) \) and \( |\vec{a}|=\frac{1}{\sqrt{2}} \), \( |\vec{b}|=\frac{1}{\sqrt{3}} \), \( |\vec{c}|=\frac{1}{\sqrt{6}} \), then the angle between \( \vec{a} \) and \( \vec{b} \) is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Hint:

Vector Tip: Always remember that $\overline{c} = \overline{a} \times \overline{b}$ implies $\overline{c}$ is orthogonal to both $\overline{a}$ and $\overline{b}$.The identity $|\overline{a}+\overline{b}+\overline{c}|^2 = \sum |\overline{a}|^2 + 2\sum (\overline{a}\cdot\overline{b})$ is extremely useful in vector geometry.

Answer 1

The Correct Option is D

Concept: Vector Algebra - Dot and Cross Products.

Step 1: Deduce dot products from the cross product condition. We are given $\overline{c}=\lambda(\overline{a}\times\overline{b})$. By definition, the cross product $\overline{a}\times\overline{b}$ is a vector perpendicular to both $\overline{a}$ and $\overline{b}$. Therefore, vector $\overline{c}$ is perpendicular to both $\overline{a}$ and $\overline{b}$. This implies their dot products are zero: $\overline{a}\cdot\overline{c} = 0$ and $\overline{b}\cdot\overline{c} = 0$.

Step 2: Expand the given magnitude equation. We are given $|\overline{a}+\overline{b}+\overline{c}|=1$. Squaring both sides gives $|\overline{a}+\overline{b}+\overline{c}|^2 = 1$. The expansion of this square is: $|\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 + 2(\overline{a}\cdot\overline{b} + \overline{b}\cdot\overline{c} + \overline{c}\cdot\overline{a}) = 1$.

Step 3: Substitute known magnitude values. We are given the magnitudes $|\overline{a}|=\frac{1}{\sqrt{2}}$, $|\overline{b}|=\frac{1}{\sqrt{3}}$, and $|\overline{c}|=\frac{1}{\sqrt{6}}$. Squaring these yields $|\overline{a}|^2 = \frac{1}{2}$, $|\overline{b}|^2 = \frac{1}{3}$, and $|\overline{c}|^2 = \frac{1}{6}$. The sum of these squares is $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$.

Step 4: Simplify the equation using the dot products. Substitute the sum of squares (which is 1) and the zero dot products (from Step 1) into the expanded equation: $1 + 2(\overline{a}\cdot\overline{b} + 0 + 0) = 1$.

This simplifies to $1 + 2(\overline{a}\cdot\overline{b}) = 1$, which means $2(\overline{a}\cdot\overline{b}) = 0$, leading to $\overline{a}\cdot\overline{b} = 0$.

Step 5: Determine the angle between the vectors. The dot product is defined as $\overline{a}\cdot\overline{b} = |\overline{a}||\overline{b}|\cos\theta$, where $\theta$ is the angle between $\overline{a}$ and $\overline{b}$. Since $\overline{a}\cdot\overline{b} = 0$ and neither magnitude is zero, it must be that $\cos\theta = 0$. Therefore, the angle $\theta$ is $\frac{\pi}{2}$. $$ \therefore \text{The angle between } \overline{a} \text{ and } \overline{b} \text{ is } \frac{\pi}{2}. $$