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Question: If \(A\equiv \left( 3,1,-2 \right);B\equiv \left( -1,0,1 \right)\) and l, m are the projections of A...

If A≡(3,1,−2);B≡(−1,0,1)A\equiv \left( 3,1,-2 \right);B\equiv \left( -1,0,1 \right) and l, m are the projections of AB on the Y-axis, Z-X plane respectively, then find the value of 3l2−m+13{{l}^{2}}-m+1.
A. -1
B. 0
C. 1
D. 9

Explanation

Solution

We try to find the direction ratios of the vector AB→\overrightarrow{AB} from the given points A≡(3,1,−2);B≡(−1,0,1)A\equiv \left( 3,1,-2 \right);B\equiv \left( -1,0,1 \right). We also form the vector in the form of projections on the axes and particular planes. We equate those values and find the solution for 3l2−m+13{{l}^{2}}-m+1.

Complete step-by-step solution:
The given points are A≡(3,1,−2);B≡(−1,0,1)A\equiv \left( 3,1,-2 \right);B\equiv \left( -1,0,1 \right).
We try to find the direction ratios of the vector AB→\overrightarrow{AB}.
So, the direction ratios will be the differences of the individual coordinates.
Therefore, AB→=(3+1,1−0,−2−1)=(4,1,−3)\overrightarrow{AB}=\left( 3+1,1-0,-2-1 \right)=\left( 4,1,-3 \right). If we express it in the form of vector, we get
AB→=4i^+j^−3k^\overrightarrow{AB}=4\widehat{i}+\widehat{j}-3\widehat{k}.
If we want to find the projection of a vector X→=axi^+ayj^+azk^\overrightarrow{X}={{a}_{x}}\widehat{i}+{{a}_{y}}\widehat{j}+{{a}_{z}}\widehat{k} on individual axes then we take the coefficients of the directions as the projection value. Here the projections on the X, Y and Z axes are ax,ay,az{{a}_{x}},{{a}_{y}},{{a}_{z}} respectively.
For projection on a plane, we take the root of the sum of the square of individual projection.
For projection on the Z-X plane, we have ax2+az2\sqrt{{{a}_{x}}^{2}+{{a}_{z}}^{2}}.
For the vector AB→=4i^+j^−3k^\overrightarrow{AB}=4\widehat{i}+\widehat{j}-3\widehat{k}, it’s given that l, m are the projections of AB on the Y-axis, Z-X plane respectively.
This means l=ay=1,m=ax2+az2=42+(−3)2=5l={{a}_{y}}=1,m=\sqrt{{{a}_{x}}^{2}+{{a}_{z}}^{2}}=\sqrt{{{4}^{2}}+{{\left( -3 \right)}^{2}}}=5.
Now we need to find the value of 3l2−m+1 3{{l}^{2}} - m+1. We place the values and get
3l2−m+1=3−5+1=−13{{l}^{2}}-m+1=3-5+1=-1. The correct option is A.

Note: We need to remember that the vector projection is the vector produced when one vector is resolved into two component vectors, one that is parallel to the 2nd vector and one that is perpendicular to the 2nd vector. The parallel vector is the vector projection. So, the coefficients remain the same for the parallel vectors and decide the projection value.