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The projection vector of the vector 2i^+3j^+k^2\hat{i} + 3\hat{j} + \hat{k} on 2i^+j^2k^2\hat{i} + \hat{j} - 2\hat{k} is

Solution

Correct Option: 2

The projection vector of a\vec{a} on b\vec{b} is given by:

Projection vector=abb2×b\text{Projection vector} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \times \vec{b}

Here, a=2i^+3j^+k^\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} and b=2i^+j^2k^\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}


Calculate the dot product ab\vec{a} \cdot \vec{b}:

ab=(2)(2)+(3)(1)+(1)(2)\vec{a} \cdot \vec{b} = (2)(2) + (3)(1) + (1)(-2)

=4+32= 4 + 3 - 2

=5= 5


Calculate b2|\vec{b}|^2:

b2=(2)2+(1)2+(2)2|\vec{b}|^2 = (2)^2 + (1)^2 + (-2)^2

=4+1+4= 4 + 1 + 4

=9= 9


Substitute into the projection formula:

Projection vector=abb2×b\text{Projection vector} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \times \vec{b}

=59×(2i^+j^2k^)= \frac{5}{9} \times (2\hat{i} + \hat{j} - 2\hat{k})

=59(2i^+j^2k^)= \frac{5}{9}(2\hat{i} + \hat{j} - 2\hat{k})

Therefore, the projection vector is 59(2i^+j^2k^)\frac{5}{9}(2\hat{i} + \hat{j} - 2\hat{k}).

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