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The projection of the vector 5i^+j^−3k^5\hat{i} + \hat{j} - 3\hat{k} on the vector i^+2j^−3k^\hat{i} + 2\hat{j} - 3\hat{k} is

Solution

✅ Correct Option: 1

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

Projection =a⃗⋅b⃗∣b⃗∣= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Given:

a⃗=5i^+j^−3k^\vec{a} = 5\hat{i} + \hat{j} - 3\hat{k}

b⃗=i^+2j^−3k^\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}


The dot product a⃗⋅b⃗\vec{a} \cdot \vec{b} is:

a⃗⋅b⃗=(5)(1)+(1)(2)+(−3)(−3)\vec{a} \cdot \vec{b} = (5)(1) + (1)(2) + (-3)(-3)

a⃗⋅b⃗=5+2+9\vec{a} \cdot \vec{b} = 5 + 2 + 9

a⃗⋅b⃗=16\vec{a} \cdot \vec{b} = 16


The magnitude of b⃗\vec{b} is:

∣b⃗∣=(1)2+(2)2+(−3)2|\vec{b}| = \sqrt{(1)^2 + (2)^2 + (-3)^2}

∣b⃗∣=1+4+9|\vec{b}| = \sqrt{1 + 4 + 9}

∣b⃗∣=14|\vec{b}| = \sqrt{14}


The projection is:

Projection =a⃗⋅b⃗∣b⃗∣= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Projection =1614= \dfrac{16}{\sqrt{14}}

Therefore, the projection of the vector 5i^+j^−3k^5\hat{i} + \hat{j} - 3\hat{k} on the vector i^+2j^−3k^\hat{i} + 2\hat{j} - 3\hat{k} is 1614\dfrac{16}{\sqrt{14}}

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