CUET UG 2025 — Mathematics Algebra

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

$$\text{Projection vector} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \times \vec{b}$$Projection vector=∣b∣2a⋅b​×b

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

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

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

$$= 4 + 3 - 2$$=4+3−2

$$= 5$$=5

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Calculate $|\vec{b}|^2$∣b∣2:

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

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

$$= 9$$=9

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Substitute into the projection formula:

$$\text{Projection vector} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \times \vec{b}$$Projection vector=∣b∣2a⋅b​×b

$$= \frac{5}{9} \times (2\hat{i} + \hat{j} - 2\hat{k})$$=95​×(2i^+j^​−2k^)

$$= \frac{5}{9}(2\hat{i} + \hat{j} - 2\hat{k})$$=95​(2i^+j^​−2k^)

Therefore, the projection vector is $\frac{5}{9}(2\hat{i} + \hat{j} - 2\hat{k})$95​(2i^+j^​−2k^).