Question

If the resolved parts of the force vector $5\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\,$ along and perpendicular to the vector $3\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,-5\overset{\wedge }{\mathop{k}}\,$ are $\alpha$ and $\beta$ respectively. Then, the value of $\alpha$ is
(A). $\dfrac{21}{50}\left( 3\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,-5\overset{\wedge }{\mathop{k}}\, \right)$
(B). $\dfrac{21}{50}\left( 3\overset{\wedge }{\mathop{i}}\,-4\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\, \right)$
(C). $\dfrac{11}{50}\left( 2\overset{\wedge }{\mathop{i}}\,-4\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right)$
(D). $\dfrac{1}{50}\left( 187\overset{\wedge }{\mathop{i}}\,+116\overset{\wedge }{\mathop{j}}\,+205\overset{\wedge }{\mathop{k}}\, \right)$

Answer 1

Hint: First break the force vector into 2 parts. Assume the force vector as F. Then assume the parallel and assume perpendicular components with 2 variables. Find the value of the parallel component in terms of F. Now apply, the projection formula on the parallel component of F and another vector. Assume another vector to be B. So the formula for projection will be applied and the result you get in this formula will be the result required. If these are 2 vectors a, b then projection of a along b will be $\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|}$.

Complete step-by-step solution -
Let us assume the force vector to be F, we can write it as:
F = $5\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\,$, $\left| \overset{\to }{\mathop{F}}\, \right|=\sqrt{{{5}^{2}}+{{4}^{2}}+{{2}^{2}}}=\sqrt{45}$
Now, take the other vector to be b, we can write it as:
b = $3\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,-5\overset{\wedge }{\mathop{k}}\,$ $\Rightarrow \left| \overset{\to }{\mathop{b}}\, \right|=\sqrt{{{3}^{2}}+{{4}^{2}}+{{5}^{2}}}=\sqrt{50}$
Now resolve the vector F into 2 parts, written as follows:
F = (parallel component) + (perpendicular component)
Let the parallel component be a. Now we need to find a.
By basic knowledge of vectors, we can say a parallel component of vector A is $A\cos \theta$.
By above condition, we can say, $a=F\cos \theta$.
So, we need the value of $\alpha$, which is asked in the question.
Given the condition of $\alpha$ in the question is written as follows:
$\alpha$ is the projection of the parallel component on a given vector.
Here, we have $\alpha$ in terms of a, b can be written as:
$\alpha$ = projection of a along b.
By basic knowledge of vector, projection formula is written as:
Projection of a along b = $\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|}$.
$\cos \theta =\dfrac{\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|}$ if $\theta$ is between $\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,$.
By substituting the value of $\overset{\to }{\mathop{a}}\,,\overset{\to }{\mathop{b}}\,$ in formula we get,
$\alpha =\dfrac{\left( F\cos \theta \right).\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|}$
By substituting the value of $\cos \theta$, we get the equation as:
$\alpha =\dfrac{\left( \overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{b}}\, \right)}{\left| \overset{\to }{\mathop{F}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|}.\dfrac{\overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{b}}\, \right|}.\left| \overset{\to }{\mathop{F}}\, \right|$
By canceling common terms, we get eh equation as:
$\alpha =\dfrac{\left( \overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{b}}\, \right)}{{{\left| \overset{\to }{\mathop{b}}\, \right|}^{2}}}=\dfrac{\left[ \left( 5\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right).\left( 3\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,-5\overset{\wedge }{\mathop{k}}\, \right) \right]}{{{\left( \sqrt{50} \right)}^{2}}}\left( 3\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,-5\overset{\wedge }{\mathop{k}}\, \right)$
By simplifying the above equation we get the value of $\alpha$ as:
$\alpha =\dfrac{21}{50}\left( 3\overset{\wedge }{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,-5\overset{\wedge }{\mathop{k}}\, \right)$
Therefore option (a) is correct for the given question.

Note: While applying projection formula be careful, which is a, which is b if you take reverse then you get $\left| a \right|$ in the denominator which is wrong. While substituting $\cos \theta$ formula see carefully that $\left| \overset{\to }{\mathop{F}}\, \right|$ will be cancelled. If you do wrong in that then answer will change look carefully at the denominator Students generally substitute $\left| \overset{\to }{\mathop{a}}\, \right|$ in place of $\left| \overset{\to }{\mathop{b}}\, \right|$ and get the wrong result. So substitute the determinant property.