Question ID: #828
Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})$. If $|\vec{a}|=1$, $|\vec{b}|=4$, $|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^\circ$ then $|\vec{a} \cdot \vec{c}|$ is:
- (1) 2
- (2) 4
- (3) 0
- (4) 1
Solution:
Rearrange the given vector equation:
$$ \vec{a} \times \vec{b} – 2(\vec{a} \times \vec{c}) = 0 $$
$$ \vec{a} \times (\vec{b} – 2\vec{c}) = 0 $$
This implies that $(\vec{b} – 2\vec{c})$ is parallel to $\vec{a}$. Let $\lambda$ be a scalar:
$$ \vec{b} – 2\vec{c} = \lambda \vec{a} $$
Squaring both sides (taking the dot product with itself):
$$ |\vec{b} – 2\vec{c}|^2 = \lambda^2 |\vec{a}|^2 $$
$$ |\vec{b}|^2 + 4|\vec{c}|^2 – 4(\vec{b} \cdot \vec{c}) = \lambda^2 |\vec{a}|^2 $$
Calculate $\vec{b} \cdot \vec{c}$ using the formula $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$:
$$ \vec{b} \cdot \vec{c} = 4 \times 2 \times \cos 60^\circ = 8 \times \frac{1}{2} = 4 $$
Substitute the values into the squared equation ($|\vec{a}|=1, |\vec{b}|=4, |\vec{c}|=2$):
$$ 16 + 4(4) – 4(4) = \lambda^2(1) $$
$$ 16 + 16 – 16 = \lambda^2 \Rightarrow \lambda^2 = 16 \Rightarrow \lambda = \pm 4 $$
Now, take the dot product of the equation $\vec{b} – 2\vec{c} = \lambda \vec{a}$ with vector $\vec{c}$:
$$ (\vec{b} – 2\vec{c}) \cdot \vec{c} = \lambda (\vec{a} \cdot \vec{c}) $$
$$ \vec{b} \cdot \vec{c} – 2|\vec{c}|^2 = \lambda (\vec{a} \cdot \vec{c}) $$
Substitute the known values:
$$ 4 – 2(4) = \lambda (\vec{a} \cdot \vec{c}) $$
$$ -4 = \lambda (\vec{a} \cdot \vec{c}) $$
$|\vec{a} \cdot \vec{c}| = 1$.
Ans. (4)
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