CUET UG 2025 — Mathematics Geometry

The line is:

$$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \mu(2\hat{i} + 3\hat{j} - \hat{k})$$r=(−i^+3j^​+k^)+μ(2i^+3j^​−k^)

From this:

• Point on the line, $A = (-1, 3, 1)$A=(−1,3,1)
• Direction ratios of the line, $\vec{d} = (2, 3, -1)$d=(2,3,−1)

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Any point $Q$Q on the line can be written as:

$$Q = (-1 + 2\mu,\ 3 + 3\mu,\ 1 - \mu)$$Q=(−1+2μ, 3+3μ, 1−μ)

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$$\vec{PQ} = Q - P$$PQ​=Q−P

$$= (-6 + 2\mu,\ -1 + 3\mu,\ -1 - \mu)$$=(−6+2μ, −1+3μ, −1−μ)

Since $\vec{PQ} \perp \vec{d}$PQ​⊥d, their dot product $= 0$=0:

$$2(-6 + 2\mu) + 3(-1 + 3\mu) + (-1)(-1 - \mu) = 0$$2(−6+2μ)+3(−1+3μ)+(−1)(−1−μ)=0

$$-12 + 4\mu - 3 + 9\mu + 1 + \mu = 0$$−12+4μ−3+9μ+1+μ=0

$$14\mu - 14 = 0$$14μ−14=0

$$\mu = 1$$μ=1

The foot of perpendicular is:

$$F = (-1 + 2(1),\ 3 + 3(1),\ 1 - 1)$$F=(−1+2(1), 3+3(1), 1−1)

$$= (1, 6, 0)$$=(1,6,0)

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Since $F$F is the midpoint of $P$P and its image $P'(x, y, z)$P′(x,y,z):

$$\dfrac{x + 5}{2} = 1 \implies x = -3$$2x+5​=1⟹x=−3

$$\dfrac{y + 4}{2} = 6 \implies y = 8$$2y+4​=6⟹y=8

$$\dfrac{z + 2}{2} = 0 \implies z = -2$$2z+2​=0⟹z=−2

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$$P' = (-3,\ 8,\ -2)$$P′=(−3, 8, −2)