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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Evaluate the product \((3\overrightarrow a − 5\overrightarrow b) ⋅ (2\overrightarrow a + 7\overrightarrow b)\)

$\begin{array}{1 1} (A) 6|\overrightarrow a|^2+35|\overrightarrow b|^2+11\overrightarrow a.\overrightarrow b \\ (B) 6|\overrightarrow a|^2-35|\overrightarrow b|^2-31\overrightarrow a.\overrightarrow b \\ (C) 6|\overrightarrow a|^2-35|\overrightarrow b|^2+31\overrightarrow a.\overrightarrow b \\ (D) 6|\overrightarrow a|^2-35|\overrightarrow b|^2+11\overrightarrow a.\overrightarrow b \end{array} $

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1 Answer

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Toolbox:
  • $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
  • $\overrightarrow a.\overrightarrow a=\mid\overrightarrow a \mid^2$
Step 1:
Given : $(3\overrightarrow a-5\overrightarrow b).(2\overrightarrow a+7\overrightarrow b)$
We are asked to find the product of the above vectors.
$(3\overrightarrow a-5\overrightarrow b).(2\overrightarrow a+7\overrightarrow b)=3\overrightarrow a(2\overrightarrow a+7\overrightarrow b)-5\overrightarrow b(2\overrightarrow a+7\overrightarrow b)$
$\qquad\qquad\qquad\qquad\quad\;\;=6\overrightarrow a.\overrightarrow a+21\overrightarrow a.\overrightarrow b-10\overrightarrow b.\overrightarrow a-35\overrightarrow b.\overrightarrow b$
Step 2:
But $\overrightarrow a.\overrightarrow a=\mid\overrightarrow a \mid^2$
$\;\;\;\;\;\overrightarrow b.\overrightarrow b=\mid\overrightarrow b \mid^2$
$\;\;\;\;\;\overrightarrow a.\overrightarrow b=\overrightarrow b. \overrightarrow a$
Therefore $(3\overrightarrow a-5\overrightarrow b).(2\overrightarrow a+7\overrightarrow b)=6|\overrightarrow a|^2-35|\overrightarrow b|^2+11\overrightarrow a.\overrightarrow b$
answered May 20, 2013 by sreemathi.v
 

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