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Question: Given the points \(A\left( -2,3,4 \right)\) , \(B\left( 3,2,5 \right)\) , \(C\left( 1,-1,2 \right)\)...

Given the points A(−2,3,4)A\left( -2,3,4 \right) , B(3,2,5)B\left( 3,2,5 \right) , C(1,−1,2)C\left( 1,-1,2 \right) and D(3,2,−4)D\left( 3,2,-4 \right) . The projection of the vector AB→\overrightarrow{AB} on the vector CD→\overrightarrow{CD} is
(A) 223\dfrac{22}{3}
(B) −214\dfrac{-21}{4}
(C) 17\dfrac{1}{7}
(D) −47-47

Explanation

Solution

For answering this question we will identify the directional rations of both the lines AB→\overrightarrow{AB} and CD→\overrightarrow{CD} . The directional ratios of a line joining (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and (x2,y2,z2)\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) is given as (x2−x1,y2−y1,z2−z1)\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}} \right) . The projection of the vector AB→\overrightarrow{AB} on the vector CD→\overrightarrow{CD} is given by the dot product of the directional ratios of both the lines divided by the length of CD→\overrightarrow{CD} .

Complete step by step answer:
Now considering from the question we have A(−2,3,4)A\left( -2,3,4 \right) , B(3,2,5)B\left( 3,2,5 \right) , C(1,−1,2)C\left( 1,-1,2 \right) and D(3,2,−4)D\left( 3,2,-4 \right) .
We know that from the basic concept that the projection of the vector AB→\overrightarrow{AB} on the vector CD→\overrightarrow{CD} is given by the dot product of the directional ratios of both the lines divided by the length of CD→\overrightarrow{CD} .
The directional ratios of a line joining (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and (x2,y2,z2)\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) is given as (x2−x1,y2−y1,z2−z1)\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}} \right) .
So here the directional ratio of AB→\overrightarrow{AB} is given as (3−(−2),2−3,5−4)=(5,−1,1)\left( 3-\left( -2 \right),2-3,5-4 \right)=\left( 5,-1,1 \right) .
The directional ratio of CD→\overrightarrow{CD} is given as (3−1,2−(−1),−4−(2))=(2,3,−6)\left( 3-1,2-\left( -1 \right),-4-\left( 2 \right) \right)=\left( 2,3,-6 \right) .
The dot product of the directional ratios of AB→\overrightarrow{AB} and CD→\overrightarrow{CD} that is(5,−1,1).(2,3,−6)=(10−3−6)=1\left( 5,-1,1 \right).\left( 2,3,-6 \right)=\left( 10-3-6 \right)=1 .
The magnitude of the line CD→\overrightarrow{CD} is given as (2)2+(3)2+(−6)2=4+9+36=49=7\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{4+9+36}=\sqrt{49}=7 .
Hence the projection of the vector AB→\overrightarrow{AB} on the vector CD→\overrightarrow{CD} is given as 17\dfrac{1}{7} .

So, the correct answer is “Option C”.

Note: While answering questions of this type we should make a note that when projecting AB→\overrightarrow{AB} on CD→\overrightarrow{CD} we should take the ratio of dot product of the directional ratios of both the lines and the length of CD→\overrightarrow{CD} . For projection of CD→\overrightarrow{CD} on AB→\overrightarrow{AB} we should take the ratio of dot product of the directional ratios of both the lines and the length of AB→\overrightarrow{AB} . If we take the other by mistake we will encounter the wrong answer.