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Question: If \(\vec a,\vec b\) and \(\vec c\) are unit vectors such that \(\vec a + 2\vec b + 2\vec c = \vec 0...

If a⃗,b⃗\vec a,\vec b and c⃗\vec c are unit vectors such that a⃗+2b⃗+2c⃗=0⃗\vec a + 2\vec b + 2\vec c = \vec 0, the ∣a⃗×c⃗∣\left| {\vec a \times \vec c} \right| is equal to:
(A) 14\dfrac{1}{4}
(B) 154\dfrac{{\sqrt {15} }}{4}
(C) 1516\dfrac{{15}}{{16}}
(D) 1516\dfrac{{\sqrt {15} }}{{16}}

Explanation

Solution

Since, a⃗,b⃗\vec a,\vec b and c⃗\vec c are unit vectors. Therefore, ∣a⃗∣=∣b⃗∣=∣c⃗∣=1\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec c} \right| = 1. Use the formulas a⃗⋅a⃗=∣a⃗∣2\vec a \cdot \vec a = {\left| {\vec a} \right|^2} and a⃗⋅b⃗=b⃗⋅a⃗\vec a \cdot \vec b = \vec b \cdot \vec a to solve the given relation a⃗+2b⃗+2c⃗=0⃗\vec a + 2\vec b + 2\vec c = \vec 0.

Complete step-by-step answer:
Given, a⃗,b⃗\vec a,\vec b and c⃗\vec c are unit vectors.
Therefore, magnitude a⃗,b⃗\vec a,\vec b and c⃗\vec c is 11.
So, ∣a⃗∣=∣b⃗∣=∣c⃗∣=1\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec c} \right| = 1
We have, a⃗+2b⃗+2c⃗=0⃗\vec a + 2\vec b + 2\vec c = \vec 0
⇒a⃗+2c⃗=−2b⃗\Rightarrow \vec a + 2\vec c = - 2\vec b
On squaring both sides,
⇒∣a⃗+2c⃗∣2=∣−2b⃗∣2\Rightarrow {\left| {\vec a + 2\vec c} \right|^2} = {\left| { - 2\vec b} \right|^2}
Using ∣a⃗∣2=a⃗⋅a⃗{\left| {\vec a} \right|^2} = \vec a \cdot \vec a,
⇒(a⃗+2c⃗)⋅(a⃗+2c⃗)=(−2b⃗)⋅(−2b⃗)\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)
⇒a⃗⋅a⃗+2a⃗⋅c⃗+2c⃗⋅a⃗+4c⃗⋅c⃗=4b⃗⋅b⃗\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 a⃗⋅a⃗=∣a⃗∣2\vec a \cdot \vec a = {\left| {\vec a} \right|^2} and a⃗⋅b⃗=b⃗⋅a⃗\vec a \cdot \vec b = \vec b \cdot \vec a,
⇒∣a⃗∣2+2a⃗⋅c⃗+2a⃗⋅c⃗+4∣c⃗∣2=4∣b⃗∣2\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}
⇒(1)2+4a⃗⋅c⃗+4(1)2=4(1)2\Rightarrow {\left( 1 \right)^2} + 4\vec a \cdot \vec c + 4{\left( 1 \right)^2} = 4{\left( 1 \right)^2}
⇒1+4a⃗⋅c⃗+4=4\Rightarrow 1 + 4\vec a \cdot \vec c + 4 = 4
⇒4a⃗⋅c⃗=−1\Rightarrow 4\vec a \cdot \vec c = - 1
⇒a⃗⋅c⃗=−14\Rightarrow \vec a \cdot \vec c = \dfrac{{ - 1}}{4}
We have, ∣a⃗⋅c⃗∣2+∣a⃗×c⃗∣2=1{\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = 1 (a⃗\vec a and c⃗\vec c are unit vectors)
⇒(−14)2+∣a⃗×c⃗∣2=1\Rightarrow {\left( {\dfrac{{ - 1}}{4}} \right)^2} + {\left| {\vec a \times \vec c} \right|^2} = 1
⇒116+∣a⃗×c⃗∣2=1\Rightarrow \dfrac{1}{{16}} + {\left| {\vec a \times \vec c} \right|^2} = 1
⇒∣a⃗×c⃗∣2=1−116\Rightarrow {\left| {\vec a \times \vec c} \right|^2} = 1 - \dfrac{1}{{16}}
⇒∣a⃗×c⃗∣2=16−116\Rightarrow {\left| {\vec a \times \vec c} \right|^2} = \dfrac{{16 - 1}}{{16}}
⇒∣a⃗×c⃗∣2=1516\Rightarrow {\left| {\vec a \times \vec c} \right|^2} = \dfrac{{15}}{{16}}
⇒∣a⃗×c⃗∣=1516\Rightarrow \left| {\vec a \times \vec c} \right| = \sqrt {\dfrac{{15}}{{16}}}
⇒∣a⃗×c⃗∣=154\Rightarrow \left| {\vec a \times \vec c} \right| = \dfrac{{\sqrt {15} }}{4}

Hence, option (B) is the correct answer.

Note: The formula ∣a⃗⋅c⃗∣2+∣a⃗×c⃗∣2=1{\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, a⃗⋅c⃗=∣a⃗∣∣c⃗∣cos⁡θ\vec a \cdot \vec c = \left| {\vec a} \right|\left| {\vec c} \right|\cos \theta
⇒∣a⃗⋅c⃗∣2=∣a⃗∣2∣c⃗∣2cos⁡2θ\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, a⃗×c⃗=∣a⃗∣∣c⃗∣sin⁡θ\vec a \times \vec c = \left| {\vec a} \right|\left| {\vec c} \right|\sin \theta
⇒∣a⃗×c⃗∣2=∣a⃗∣2∣c⃗∣2sin⁡2θ\Rightarrow {\left| {\vec a \times \vec c} \right|^2} = {\left| {\vec a} \right|^2}{\left| {\vec c} \right|^2}{\sin ^2}\theta
Now, ∣a⃗⋅c⃗∣2+∣a⃗×c⃗∣2=∣a⃗∣2∣c⃗∣2cos⁡2θ+∣a⃗∣2∣c⃗∣2sin⁡2θ{\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
⇒∣a⃗⋅c⃗∣2+∣a⃗×c⃗∣2=∣a⃗∣2∣c⃗∣2(cos⁡2θ+sin⁡2θ)\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)
⇒∣a⃗⋅c⃗∣2+∣a⃗×c⃗∣2=(1)2(1)2(1)\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)
(∵sin⁡2θ+cos⁡2θ=1)\left( {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right)
⇒∣a⃗⋅c⃗∣2+∣a⃗×c⃗∣2=1\Rightarrow {\left| {\vec a \cdot \vec c} \right|^2} + {\left| {\vec a \times \vec c} \right|^2} = 1