CUET UG 2025 — Mathematics Arithmetic

$$\vec{r}_1 = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$$r1​=(i^+2j^​+3k^)+λ(2i^+3j^​+4k^)

$$\vec{r}_2 = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$$r2​=(2i^+4j^​+5k^)+μ(3i^+4j^​+5k^)

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$$\vec{a}_1 = (1, 2, 3), \quad \vec{b}_1 = (2, 3, 4)$$a1​=(1,2,3),b1​=(2,3,4)

$$\vec{a}_2 = (2, 4, 5), \quad \vec{b}_2 = (3, 4, 5)$$a2​=(2,4,5),b2​=(3,4,5)

The shortest distance formula:

$$d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$d=∣b1​×b2​∣∣(a2​−a1​)⋅(b1​×b2​)∣​

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$$\vec{a}_2 - \vec{a}_1 = (1, 2, 2)$$a2​−a1​=(1,2,2)

$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$$b1​×b2​=​i^23​j^​34​k^45​​

$$= (-1)\hat{i} + (2)\hat{j} + (-1)\hat{k}$$=(−1)i^+(2)j^​+(−1)k^

$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1 + 4 + 1} = \sqrt{6}$$∣b1​×b2​∣=1+4+1​=6​

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$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-1) + (2)(2) + (2)(-1)$$(a2​−a1​)⋅(b1​×b2​)=(1)(−1)+(2)(2)+(2)(−1)

$$= -1 + 4 - 2 = 1$$=−1+4−2=1

$$d = \dfrac{|1|}{\sqrt{6}}$$d=6​∣1∣​

$$\boxed{d = \dfrac{1}{\sqrt{6}}}$$d=6​1​​