CUET UG 2025 — Mathematics Algebra

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

Projection $= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$=∣b∣a⋅b​

Given:

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

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

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

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

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

$$\vec{a} \cdot \vec{b} = 16$$a⋅b=16

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The magnitude of $\vec{b}$b is:

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

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

$$|\vec{b}| = \sqrt{14}$$∣b∣=14​

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The projection is:

Projection $= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$=∣b∣a⋅b​

Projection $= \dfrac{16}{\sqrt{14}}$=14​16​

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