Question

If $\vec a,\vec b$ and $\vec c$ are unit vectors such that $\vec a + 2\vec b + 2\vec c = \vec 0$, the $\left| {\vec a \times \vec c} \right|$ is equal to:
(A) $\dfrac{1}{4}$
(B) $\dfrac{{\sqrt {15} }}{4}$
(C) $\dfrac{{15}}{{16}}$
(D) $\dfrac{{\sqrt {15} }}{{16}}$

Answer 1

Hint: Since, $\vec a,\vec b$ and $\vec c$ are unit vectors. Therefore, $\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec c} \right| = 1$. Use the formulas \[\vec a \cdot \vec a = {\left| {\vec a} \right|^2}\] and \[\vec a \cdot \vec b = \vec b \cdot \vec a\] to solve the given relation $\vec a + 2\vec b + 2\vec c = \vec 0$.

Complete step-by-step answer:
Given, $\vec a,\vec b$ and $\vec c$ are unit vectors.
Therefore, magnitude $\vec a,\vec b$ and $\vec c$ is $1$.
So, $\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec c} \right| = 1$
We have, $\vec a + 2\vec b + 2\vec c = \vec 0$
$ \Rightarrow \vec a + 2\vec c = - 2\vec b$
On squaring both sides,
$ \Rightarrow {\left| {\vec a + 2\vec c} \right|^2} = {\left| { - 2\vec b} \right|^2}$
Using \[{\left| {\vec a} \right|^2} = \vec a \cdot \vec a\],
$ \Rightarrow \left( {\vec a + 2\vec c} \right) \cdot \left( {\vec a + 2\vec c} \right) = \left( { - 2\vec b} \right) \cdot \left( { - 2\vec b} \right)$
\[ \Rightarrow \vec a \cdot \vec a + 2\vec a \cdot \vec c + 2\vec c \cdot \vec a + 4\vec c \cdot \vec c = 4\vec b \cdot \vec b\]
Now using \[\vec a \cdot \vec a = {\left| {\vec a} \right|^2}\] and \[\vec a \cdot \vec b = \vec b \cdot \vec a\],
\[ \Rightarrow {\left| {\vec a} \right|^2} + 2\vec a \cdot \vec c + 2\vec a \cdot \vec c + 4{\left| {\vec c} \right|^2} = 4{\left| {\vec b} \right|^2}\]
\[ \Rightarrow {\left( 1 \right)^2} + 4\vec a \cdot \vec c + 4{\left( 1 \right)^2} = 4{\left( 1 \right)^2}\]
\[ \Rightarrow 1 + 4\vec a \cdot \vec c + 4 = 4\]
\[ \Rightarrow 4\vec a \cdot \vec c = - 1\]
\[ \Rightarrow \vec a \cdot \vec c = \dfrac{{ - 1}}{4}\]
We have, ${\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = 1$ ($\vec a$ and $\vec c$ are unit vectors)
$ \Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^2} + {\left| {\vec a \times \vec c} \right|^2} = 1$
$ \Rightarrow \dfrac{1}{{16}} + {\left| {\vec a \times \vec c} \right|^2} = 1$
$ \Rightarrow {\left| {\vec a \times \vec c} \right|^2} = 1 - \dfrac{1}{{16}}$
$ \Rightarrow {\left| {\vec a \times \vec c} \right|^2} = \dfrac{{16 - 1}}{{16}}$
$ \Rightarrow {\left| {\vec a \times \vec c} \right|^2} = \dfrac{{15}}{{16}}$
\[ \Rightarrow \left| {\vec a \times \vec c} \right| = \sqrt {\dfrac{{15}}{{16}}} \]
\[ \Rightarrow \left| {\vec a \times \vec c} \right| = \dfrac{{\sqrt {15} }}{4}\]

Hence, option (B) is the correct answer.

Note: The formula ${\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = 1$ can be derived as follows:
We know that, $\vec a \cdot \vec c = \left| {\vec a} \right|\left| {\vec c} \right|\cos \theta $
$ \Rightarrow {\left| {\vec a \cdot \vec c} \right|^2} = {\left| {\vec a} \right|^2}{\left| {\vec c} \right|^2}{\cos ^2}\theta $
We also know that, $\vec a \times \vec c = \left| {\vec a} \right|\left| {\vec c} \right|\sin \theta $
$ \Rightarrow {\left| {\vec a \times \vec c} \right|^2} = {\left| {\vec a} \right|^2}{\left| {\vec c} \right|^2}{\sin ^2}\theta $
Now, ${\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = {\left| {\vec a} \right|^2}{\left| {\vec c} \right|^2}{\cos ^2}\theta + {\left| {\vec a} \right|^2}{\left| {\vec c} \right|^2}{\sin ^2}\theta $
$ \Rightarrow {\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = {\left| {\vec a} \right|^2}{\left| {\vec c} \right|^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)$
$ \Rightarrow {\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = {\left( 1 \right)^2}{\left( 1 \right)^2}\left( 1 \right)$
$\left( {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right)$
$ \Rightarrow {\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = 1$