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Question: What will be the projection of vector $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$ on vector $...

What will be the projection of vector A→=i^+j^+k^\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k} on vector B→=i^+j^\overrightarrow{B} = \hat{i} + \hat{j} ? [JEE-Main-2021_July]

A

2(i^+j^+k^)\sqrt{2}(\hat{i} + \hat{j} + \hat{k})

B

2(i^+j^+k^)2(\hat{i} + \hat{j} + \hat{k})

C

2(i^+j^)\sqrt{2}(\hat{i} + \hat{j})

D

(i^+j^)(\hat{i} + \hat{j})

Answer

i^+j^\hat{i} + \hat{j}

Explanation

Solution

The projection of vector A→\overrightarrow{A} on vector B→\overrightarrow{B} is given by the formula: projB→A→=A→⋅B→∣∣B→∣∣2B→\text{proj}_{\overrightarrow{B}} \overrightarrow{A} = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{||\overrightarrow{B}||^2} \overrightarrow{B} Given vectors are A→=i^+j^+k^\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k} and B→=i^+j^\overrightarrow{B} = \hat{i} + \hat{j}.

First, calculate the dot product A→⋅B→\overrightarrow{A} \cdot \overrightarrow{B}: A→⋅B→=(i^+j^+k^)⋅(i^+j^)=(1)(1)+(1)(1)+(1)(0)=1+1+0=2\overrightarrow{A} \cdot \overrightarrow{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j}) = (1)(1) + (1)(1) + (1)(0) = 1 + 1 + 0 = 2

Next, calculate the magnitude of vector B→\overrightarrow{B}, ∣∣B→∣∣||\overrightarrow{B}||: ∣∣B→∣∣=(1)2+(1)2+(0)2=1+1+0=2||\overrightarrow{B}|| = \sqrt{(1)^2 + (1)^2 + (0)^2} = \sqrt{1 + 1 + 0} = \sqrt{2}

Now, calculate the square of the magnitude of vector B→\overrightarrow{B}, ∣∣B→∣∣2||\overrightarrow{B}||^2: ∣∣B→∣∣2=(2)2=2||\overrightarrow{B}||^2 = (\sqrt{2})^2 = 2

Substitute the calculated values into the projection formula: projB→A→=22(i^+j^)=1⋅(i^+j^)=i^+j^\text{proj}_{\overrightarrow{B}} \overrightarrow{A} = \frac{2}{2} (\hat{i} + \hat{j}) = 1 \cdot (\hat{i} + \hat{j}) = \hat{i} + \hat{j}