JEE Main 2026 — Mathematics Vectors & 3D Geometry

Given $2(\vec{a}\times\vec{b}) + 3(\vec{b}\times\vec{c}) = \vec{0}$2(a×b)+3(b×c)=0

$$\Rightarrow 2(\vec{a}\times\vec{b}) - 3(\vec{c}\times\vec{b}) = \vec{0}$$⇒2(a×b)−3(c×b)=0

$$\Rightarrow (2\vec{a} - 3\vec{c}) \times \vec{b} = \vec{0}$$⇒(2a−3c)×b=0

Since the cross product is zero, the vectors are collinear:

$$2\vec{a} - 3\vec{c} = \lambda \vec{b}$$2a−3c=λb

$$\Rightarrow 3\vec{c} = 2\vec{a} - \lambda \vec{b}$$⇒3c=2a−λb

Taking the dot product with $\vec{a}$a on both sides:

$$3(\vec{a}\cdot\vec{c}) = 2|\vec{a}|^2 - \lambda(\vec{a}\cdot\vec{b})$$3(a⋅c)=2∣a∣2−λ(a⋅b)

We have $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$a=4i^−j^​+3k^ and $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$b=10i^+2j^​−k^.

$$|\vec{a}|^2 = 4^2 + (-1)^2 + 3^2 = 26$$∣a∣2=42+(−1)2+32=26

$$\vec{a}\cdot\vec{b} = 4(10) + (-1)(2) + 3(-1) = 40 - 2 - 3 = 35$$a⋅b=4(10)+(−1)(2)+3(−1)=40−2−3=35

Given $\vec{a}\cdot\vec{c} = 15$a⋅c=15, substituting these values:

$$3(15) = 2(26) - 35\lambda$$3(15)=2(26)−35λ

$$45 = 52 - 35\lambda$$45=52−35λ

$$35\lambda = 7 \Rightarrow \lambda = \dfrac{1}{5}$$35λ=7⇒λ=51​

Substituting $\lambda$λ back into the equation for $\vec{c}$c:

$$3\vec{c} = 2\vec{a} - \dfrac{1}{5}\vec{b}$$3c=2a−51​b

We need to find $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$c⋅(i^+j^​−3k^). Let $\vec{v} = \hat{i}+\hat{j}-3\hat{k}$v=i^+j^​−3k^.

Taking the dot product with $\vec{v}$v on both sides:

$$3(\vec{c}\cdot\vec{v}) = 2(\vec{a}\cdot\vec{v}) - \dfrac{1}{5}(\vec{b}\cdot\vec{v})$$3(c⋅v)=2(a⋅v)−51​(b⋅v)

Calculating $\vec{a}\cdot\vec{v}$a⋅v and $\vec{b}\cdot\vec{v}$b⋅v:

$$\vec{a}\cdot\vec{v} = 4(1) + (-1)(1) + 3(-3) = 4 - 1 - 9 = -6$$a⋅v=4(1)+(−1)(1)+3(−3)=4−1−9=−6

$$\vec{b}\cdot\vec{v} = 10(1) + 2(1) + (-1)(-3) = 10 + 2 + 3 = 15$$b⋅v=10(1)+2(1)+(−1)(−3)=10+2+3=15

Substituting these values:

$$3(\vec{c}\cdot\vec{v}) = 2(-6) - \dfrac{1}{5}(15)$$3(c⋅v)=2(−6)−51​(15)

$$3(\vec{c}\cdot\vec{v}) = -12 - 3 = -15$$3(c⋅v)=−12−3=−15

$$\vec{c}\cdot\vec{v} = -5$$c⋅v=−5

Answer: $-5$−5