Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 14th Jun 2024
Total views: 392.7k
Views today: 7.92k
Verified
392.7k+ views
Hint: For answering this question we will identify the directional rations of both the lines $\overrightarrow{AB}$ and $\overrightarrow{CD}$ . The directional ratios of a line joining $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given as $\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}} \right)$ . The projection of the vector $\overrightarrow{AB}$ on the vector $\overrightarrow{CD}$ is given by the dot product of the directional ratios of both the lines divided by the length of $\overrightarrow{CD}$ .

Now considering from the question we have $A\left( -2,3,4 \right)$ , $B\left( 3,2,5 \right)$ , $C\left( 1,-1,2 \right)$ and $D\left( 3,2,-4 \right)$ .
We know that from the basic concept that the projection of the vector $\overrightarrow{AB}$ on the vector $\overrightarrow{CD}$ is given by the dot product of the directional ratios of both the lines divided by the length of $\overrightarrow{CD}$ .
The directional ratios of a line joining $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given as $\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}} \right)$ .
So here the directional ratio of $\overrightarrow{AB}$ is given as $\left( 3-\left( -2 \right),2-3,5-4 \right)=\left( 5,-1,1 \right)$ .
The directional ratio of $\overrightarrow{CD}$ is given as $\left( 3-1,2-\left( -1 \right),-4-\left( 2 \right) \right)=\left( 2,3,-6 \right)$ .
The dot product of the directional ratios of $\overrightarrow{AB}$ and $\overrightarrow{CD}$ that is$\left( 5,-1,1 \right).\left( 2,3,-6 \right)=\left( 10-3-6 \right)=1$ .
The magnitude of the line $\overrightarrow{CD}$ is given as $\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{4+9+36}=\sqrt{49}=7$ .
Hence the projection of the vector $\overrightarrow{AB}$ on the vector $\overrightarrow{CD}$ is given as $\dfrac{1}{7}$ .
Note: While answering questions of this type we should make a note that when projecting $\overrightarrow{AB}$ on $\overrightarrow{CD}$ we should take the ratio of dot product of the directional ratios of both the lines and the length of $\overrightarrow{CD}$ . For projection of $\overrightarrow{CD}$ on $\overrightarrow{AB}$ we should take the ratio of dot product of the directional ratios of both the lines and the length of $\overrightarrow{AB}$ . If we take the other by mistake we will encounter the wrong answer.