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# How do you find the scalar and vector projections of $b$ onto a given $a = < 3,\; - 6,\;2 > ,\;b = < 1,\;1,\;1 > ?$

Last updated date: 12th Aug 2024
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Hint:Scalar projection of $b$ onto $a$ is given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{\left| a \right|}}$ and vector projection of $b$ onto $a$ is given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{{{\left( {\left| a \right|} \right)}^2}}}\overrightarrow a$
Where $\overrightarrow a .\;\overrightarrow b$ represents the dot product between $a\;{\text{and}}\;b$, $\left| a \right|$ represents the magnitude of the vector $a$.
Use the above information to find the respective scalar and vector projections.

In order to find the scalar projection of $b$ onto $a$ we have to first find the dot product of $a\;{\text{and}}\;b$. Dot product of two vectors $x = < a,\;b,\;c > \;{\text{and}}\;y = < e,\;f,\;g >$ is given as follows
$\overrightarrow x .\overrightarrow y = < a,\;b,\;c > . < e,\;f,\;g > = (a \times e + b \times f + c \times g)$
From the above formula of dot product performing the dot product of $a\;{\text{and}}\;b$, we will get,
$\overrightarrow a .\;\overrightarrow b = < 3,\; - 6,\;2 > . < 1,\;1,\;1 > = \left( {3 \times 1 + ( - 6) \times 1 + 2 \times 1} \right) = 3 - 6 + 2 = - 1 \\ \Rightarrow \overrightarrow a .\;\overrightarrow b = - 1 \\$
Now we have to find the magnitude of $a$ in order to find the scalar projection of $b$ onto $a$. Magnitude of a vector $x = < a,\;b,\;c >$ is given as
$\left| x \right| = \left| {\sqrt {{a^2} + {b^2} + {c^2}} } \right|$
Using this to find the magnitude of $a = < 3,\; - 6,\;2 >$, we will get
$\left| a \right| = \left| {\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} } \right| = \left| {\sqrt {9 + 36 + 4} } \right| = \left| {\sqrt {49} } \right| = \left| { \pm 7} \right| = 7$
Scalar projection of a vector $x = < a,\;b,\;c >$ on to the vector $y = < e,\;f,\;g >$ is given as $\dfrac{{\overrightarrow y .\;\overrightarrow x }}{{\left| y \right|}}$
With the use of this formula the scalar projection of $b$ onto $a$ will be given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{\left| a \right|}}$
Putting the values we will get
$\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{\left| a \right|}} = \dfrac{{ - 1}}{7}$
Now we will find the vector projection of $b$ onto $a$
Vector projection of a vector $x = < a,\;b,\;c >$ on to the vector $y = < e,\;f,\;g >$ is given as $\dfrac{{\overrightarrow y .\;\overrightarrow x }}{{{{\left( {\left| y \right|} \right)}^2}}}\overrightarrow y$
Therefore vector projection of $b$ onto $a$ will be given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{{{\left( {\left| a \right|} \right)}^2}}}\overrightarrow a$
Hence we are familiar to these terms in the vector projection of $b$ onto $a$
So directly putting their values we will get
$\therefore\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{{{\left( {\left| a \right|} \right)}^2}}}\overrightarrow a = \dfrac{{ - 1}}{{{7^2}}} < 3,\; - 6,\;2 > = \dfrac{{ - 1}}{{49}} < 3,\; - 6,\;2 >$

Therefore the required scalar and vector projections of the given vectors are $\dfrac{{ - 1}}{7}\;{\text{and}}\;\dfrac{{ - 1}}{{49}} < 3,\; - 6,\;2 >$ respectively.

Note:The given vectors $a = < 3,\; - 6,\;2 > \;{\text{and}}\;b = < 1,\;1,\;1 >$ can also be written as $a = \left( {3\widehat {\text{i}},\; - 6\widehat {\text{j}},\;2\widehat {\text{k}}} \right)\;{\text{and}}\;b = \left( {\widehat {\text{i}},\;\widehat {\text{j}},\;\widehat {\text{k}}} \right)$ where it $3\widehat {\text{i}}$ is read as “3 i-cap” it is also a form vector notation. Also the actual formula for dot product of two vectors $\overrightarrow x \;{\text{and}}\;\overrightarrow y$ is given as follows
$\overrightarrow x .\overrightarrow y = xy\cos \theta ,\;{\text{where}}\;\theta$ is the acute angle between the two vectors.