Let $\overrightarrow a$, $\overrightarrow b$ and $\overrightarrow c$ be three unit vectors such that
${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$ = 8.
Then ${\left| {\overrightarrow a + 2\overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a + 2\overrightarrow c } \right|^2}$ is equal to ______.
Answer (integer)
2
Solution
Given, $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = 1$$
<br><br>${\left| {\overrightarrow a - \overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a - \overrightarrow c } \right|^2}$ = 8
<br><br>$\Rightarrow$ $${\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} - 2\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow c } \right|^2} - 2\overrightarrow a .\overrightarrow c $$ = 8
<br><br>$\Rightarrow$ $\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c$ = -2
<br><br>Now, ${\left| {\overrightarrow a + 2\overrightarrow b } \right|^2}$ + ${\left| {\overrightarrow a + 2\overrightarrow c } \right|^2}$
<br><br>= $${\left| {\overrightarrow a } \right|^2} + 4{\left| {\overrightarrow b } \right|^2} + 4\overrightarrow a .\overrightarrow b + {\left| {\overrightarrow a } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 4\overrightarrow a .\overrightarrow c $$
<br><br>= 10 + 4$$\left( {\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c } \right)$$
<br><br>= 10 + 4(-2)
<br><br>= 2
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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