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Question: Let \(P=\left( 1,0,-1 \right)\), \(Q=\left( -1,2,0 \right)\), \(R=\left( 2,0,-3 \right)\)and \(S=\le...

Let P=(1,0,−1)P=\left( 1,0,-1 \right), Q=(−1,2,0)Q=\left( -1,2,0 \right), R=(2,0,−3)R=\left( 2,0,-3 \right)and S=(3,−2,−1)S=\left( 3,-2,-1 \right). Then the length of the components of RS on PQ is?

Explanation

Solution

If P, Q, R, and S are four vector point then the component RS on PQ is ∣RS→.PQ→∣PQ→∣∣\left| \dfrac{\overrightarrow{RS}.\overrightarrow{PQ}}{\left| \overrightarrow{PQ} \right|} \right|.
The vector from one fixed point to another point is called the position vector.
The magnitude is the distance from the starting point to the endpoint of the vector is represented by ′′∣∣′′''\left| {} \right|''.
The vector projection P on Q is the length of segment PQ.
Let’s the starting point in P and the endpoint Q then the magnitude of PQ will be
⇒∣PQ→∣=(x1−x2)2+(y1−y2)2+(z1−z2)2\Rightarrow \left| \overrightarrow{PQ} \right|=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}

Complete step by step solution:
The vector points are P=(1,0,−1)P=\left( 1,0,-1 \right), Q=(−1,2,0)Q=\left( -1,2,0 \right), R=(2,0,−3)R=\left( 2,0,-3 \right)and S=(3,−2,−1)S=\left( 3,-2,-1 \right).
The position vector OP→\overrightarrow{OP}for a point P=(1,0,−1)P=(1,0,-1)is
⇒OP→=1.i^+0.j^−1.k^\Rightarrow \overrightarrow{OP}=1.\hat{i}+0.\hat{j}-1.\hat{k}
⇒OP→=i^−k^\Rightarrow \overrightarrow{OP}=\hat{i}-\hat{k}
Similarly, the position vector for points Q, R, and S will be
⇒OQ→=−i^+2j^\Rightarrow \overrightarrow{OQ}=-\hat{i}+2\hat{j}
⇒OR→=2i^−3k^\Rightarrow \overrightarrow{OR}=2\hat{i}-3\hat{k}
⇒OS→=3i^−2j^−k^\Rightarrow \overrightarrow{OS}=3\hat{i}-2\hat{j}-\hat{k}
Where, O is the origin point (0,0,0)(0,0,0).
Now the component PQ→\overrightarrow{PQ}is equal to OQ→−OP→\overrightarrow{OQ}-\overrightarrow{OP}, therefore the component
⇒PQ→=OQ→−OP→\Rightarrow \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}
⇒PQ→=−i^+2j^−(i^−k^)=−2i^+2j^+k^\Rightarrow \overrightarrow{PQ}=-\hat{i}+2\hat{j}-\left( \hat{i}-\hat{k} \right)=-2\hat{i}+2\hat{j}+\hat{k}
⇒PQ→=−2i^+2j^+k^\Rightarrow \overrightarrow{PQ}=-2\hat{i}+2\hat{j}+\hat{k}
Similarly, the component RS→\overrightarrow{RS}is
⇒RS→=OS→−OR→=i^−2j^+2k^\Rightarrow \overrightarrow{RS}=\overrightarrow{OS}-\overrightarrow{OR}=\hat{i}-2\hat{j}+2\hat{k}
⇒RS→=i^−2j^+2k^\Rightarrow \overrightarrow{RS}=\hat{i}-2\hat{j}+2\hat{k}
Thus, the length of the components RS→\overrightarrow{RS}on PQ→\overrightarrow{PQ} =∣PQ→.RS→∣PQ→∣∣=\left| \dfrac{\overrightarrow{PQ}.\overrightarrow{RS}}{\left| \overrightarrow{PQ} \right|} \right|
The magnitude of ∣PQ→∣=x2+y2+z2\left| \overrightarrow{PQ} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}
⇒∣PQ→∣=(−2)2+22+12\Rightarrow \left| \overrightarrow{PQ} \right|=\sqrt{{{(-2)}^{2}}+{{2}^{2}}+{{1}^{2}}}
⇒∣PQ→∣=4+4+1=9\Rightarrow \left| \overrightarrow{PQ} \right|=\sqrt{4+4+1}=\sqrt{9}
⇒∣PQ→∣=3\Rightarrow \left| \overrightarrow{PQ} \right|=3
So, c PQ→=∣(i^−2j^+2k^).(−2i^+2j^+k^)3∣\overrightarrow{PQ}=\left| \dfrac{\left( \hat{i}-2\hat{j}+2\hat{k} \right).\left( -2\hat{i}+2\hat{j}+\hat{k} \right)}{3} \right|
⇒∣−2−4+23∣=∣−43∣\Rightarrow \left| \dfrac{-2-4+2}{3} \right|=\left| \dfrac{-4}{3} \right|
The components RS→\overrightarrow{RS}on PQ→=∣−43∣=43\overrightarrow{PQ}=\left| \dfrac{-4}{3} \right|=\dfrac{4}{3}

Hence, the components RS→\overrightarrow{RS} on PQ→\overrightarrow{PQ}is 43\dfrac{4}{3}.

Note:
P and Q or two vector quantities then the dot product of both vector quantity will be scalar.
It is denoted by ′(.)′'(.)'the symbol dot. For example, P and Q are two vectors then the dot product will be
⇒P→.Q→=∣P→∣∣Q→∣cos⁡θ\Rightarrow \overrightarrow{P}.\overrightarrow{Q}=\left| \overrightarrow{P} \right|\left| \overrightarrow{Q} \right|\cos \theta where θ\theta is the angle between two vectors P and Q.