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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Projection vector of $\overrightarrow{a}$ on $\overrightarrow{b}$ is

\[(A)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{b}\mid^2}.\overrightarrow{b}\quad(B)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{b}\mid}\quad(C)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{a}\mid}\quad(D)\;\frac{\overrightarrow{a}.\overrightarrow{b}}{\mid \overrightarrow{a}\mid^2}.\overrightarrow{b}\]
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  • Projection of $\overrightarrow a$ on $\overrightarrow b$ is $\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|} $$\hat b$
Projection of $\overrightarrow a$ on $\overrightarrow b$ is $\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|^2} $$\overrightarrow b $
If geometically interpreted, the scalar product of two vectors in the product of modulus of either vector and the projection of the other in its direction.
Thus, the projection of $\overrightarrow a$ on $\overrightarrow b$ is the dot product of $\overrightarrow a$ with the unit vector along $ \overrightarrow b$. is $\large\frac{\overrightarrow b}{|\overrightarrow b|}$
Therefore projection of $\overrightarrow a$ on $\overrightarrow b$ is $\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|^2}$$\:\hat b$
$=\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|^2}$$\:\overrightarrow b$
Hene the correct option is $A$
answered May 29, 2013 by meena.p
 

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