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Vector Algebra question

2012 · Shift 1 · Q23
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  5. /2012 · Shift 1 · Q23

Vector Algebra question

2012 · Shift 1 · Q23

JEE AdvancedMathematicsVector AlgebraNumerical+4 / −1
If a→,b→\overrightarrow a ,\overrightarrow ba,b and c→\overrightarrow cc are unit vectors satisfying ∣a→−b→∣2+∣b→−c→∣2+∣c→−a→∣2=9,{\left| {\overrightarrow a - \overrightarrow b } \right|^2} + {\left| {\overrightarrow b - \overrightarrow c } \right|^2} + {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = 9,​a−b​2+​b−c​2+​c−a​2=9, then ∣2a→+5b→+5c→∣\left| {2\overrightarrow a + 5\overrightarrow b + 5\overrightarrow c } \right|​2a+5b+5c​ is
Numerical answer
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Correct answer: 3

  1. We use the identity
dot vec y.$$ Since $\vec a,\vec b,\vec c$ are unit vectors, $$|\vec a|^2=|\vec b|^2=|\vec c|^2=1.$$ 2. Expand each term: $$|\vec a-\vec b|^2=1+1-2\vec a\cdot\vec b=2-2\vec a\cdot\vec b,$$ $$|\vec b-\vec c|^2=2-2\vec b\cdot\vec c,$$ $$|\vec c-\vec a|^2=2-2\vec c\cdot\vec a.$$ So, $$|\vec a-\vec b|^2+|\vec b-\vec c|^2+|\vec c-\vec a|^2 =6-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a).$$ Given this equals $9$, we get $$6-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=9.$$ Hence $$-2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=3,$$ $$\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a=-\frac32.$$ 3. Now compute $$|2\vec a+5\vec b+5\vec c|^2.$$ Using $|\vec u|^2=\vec u\cdot\vec u$, \begin{align*} |2\vec a+5\vec b+5\vec c|^2 &=(2\vec a+5\vec b+5\vec c)\cdot(2\vec a+5\vec b+5\vec c)\\ &=4|\vec a|^2+25|\vec b|^2+25|\vec c|^2+20\vec a\cdot\vec b+20\vec a\cdot\vec c+50\vec b\cdot\vec c. \end{align*} Since all are unit vectors, $$=4+25+25+20(\vec a\cdot\vec b+\vec a\cdot\vec c)+50\vec b\cdot\vec c.$$ This does not simplify directly from only the sum of pairwise dot products, so rewrite cleverly: $$2\vec a+5\vec b+5\vec c=2(\vec a+\vec b+\vec c)+3(\vec b+\vec c).$$ A better route is to use the given condition to determine the configuration. 4. Note that for unit vectors, $$|\vec a-\vec b|^2, |\vec b-\vec c|^2, |\vec c-\vec a|^2\le 4.$$ Their sum is $9$. Also, $$|\vec a+\vec b+\vec c|^2=|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a).$$ Substitute the value found: $$|\vec a+\vec b+\vec c|^2=3+2\left(-\frac32\right)=0.$$ Therefore, $$\vec a+\vec b+\vec c=\vec 0.$$ So, $$\vec b+\vec c=-\vec a.$$ 5. Now evaluate the required vector: $$2\vec a+5\vec b+5\vec c=2\vec a+5(\vec b+\vec c)=2\vec a+5(-\vec a)=-3\vec a.$$ Therefore, $$|2\vec a+5\vec b+5\vec c|=|-3\vec a|=3|\vec a|=3.$$ 6. Final answer: $$\boxed{3}$$
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