CUET UG 2025 — Mathematics Calculus

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

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

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Here, $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$a=i^+2j^​−3k^ and $\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$b=2i^+6j^​+3k^

The dot product is found by multiplying matching components and adding them up:

$$\vec{a} \cdot \vec{b} = (1)(2) + (2)(6) + (-3)(3)$$a⋅b=(1)(2)+(2)(6)+(−3)(3)

$$= 2 + 12 - 9$$=2+12−9

$$= 5$$=5

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

$$|\vec{b}| = \sqrt{(2)^2 + (6)^2 + (3)^2}$$∣b∣=(2)2+(6)2+(3)2​

$$= \sqrt{4 + 36 + 9}$$=4+36+9​

$$= \sqrt{49}$$=49​

$$= 7$$=7

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Substituting into the formula:

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

$$= \dfrac{5}{7}$$=75​

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The projection of $\vec{a}$a on $\vec{b}$b is

$$\boxed{\dfrac{5}{7}}$$75​​