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MathMedMain 2026 • 8th Apr - Shift 2

Calculate Dot Product of Vector C with Given Unit Vector Combination

Let a⃗=4i^−j^+3k^\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}, b⃗=10i^+2j^−k^\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k} and a vector c⃗\vec{c} be such that 2(a⃗×b⃗)+3(b⃗×c⃗)=0⃗2\left(\vec{a} \times \vec{b}\right) + 3\left(\vec{b} \times \vec{c}\right) = \vec{0}. If a⃗⋅c⃗=15\vec{a} \cdot \vec{c} = 15, then c⃗⋅(i^+j^−3k^)\vec{c} \cdot \left(\hat{i} + \hat{j} - 3\hat{k}\right) is equal to :

Options

A

−6-6

B

−5-5

Correct
C

−4-4

D

−3-3

Topics & Concepts

Vector AlgebraDot and Cross ProductVector Addition

Step-by-Step Solution

To find the value of c⃗⋅(i^+j^−3k^)\vec{c} \cdot \left(\hat{i} + \hat{j} - 3\hat{k}\right), we start with the given vector equation:

2(a⃗×b⃗)+3(b⃗×c⃗)=0⃗2\left(\vec{a} \times \vec{b}\right) + 3\left(\vec{b} \times \vec{c}\right) = \vec{0}

Using the property of the cross product a⃗×b⃗=−(b⃗×a⃗)\vec{a} \times \vec{b} = -\left(\vec{b} \times \vec{a}\right), we rewrite the equation as:

−2(b⃗×a⃗)+3(b⃗×c⃗)=0⃗-2\left(\vec{b} \times \vec{a}\right) + 3\left(\vec{b} \times \vec{c}\right) = \vec{0}

b⃗×(3c⃗−2a⃗)=0⃗\vec{b} \times \left(3\vec{c} - 2\vec{a}\right) = \vec{0}

Since the cross product of b⃗\vec{b} and (3c⃗−2a⃗)\left(3\vec{c} - 2\vec{a}\right) is the zero vector, 3c⃗−2a⃗3\vec{c} - 2\vec{a} must be parallel to b⃗\vec{b}. Thus, there exists a scalar λ∈R\lambda \in \mathbb{R} such that:

3c⃗−2a⃗=λb⃗  ⟹  c⃗=13(2a⃗+λb⃗)3\vec{c} - 2\vec{a} = \lambda \vec{b} \implies \vec{c} = \frac{1}{3}\left(2\vec{a} + \lambda \vec{b}\right)

Next, taking the dot product of both sides with a⃗\vec{a}:

a⃗⋅c⃗=13(2(a⃗⋅a⃗)+λ(a⃗⋅b⃗))\vec{a} \cdot \vec{c} = \frac{1}{3}\left(2\left(\vec{a} \cdot \vec{a}\right) + \lambda\left(\vec{a} \cdot \vec{b}\right)\right)

We are given a⃗=4i^−j^+3k^\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k} and b⃗=10i^+2j^−k^\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}. Let's compute a⃗⋅a⃗\vec{a} \cdot \vec{a} and a⃗⋅b⃗\vec{a} \cdot \vec{b}:

a⃗⋅a⃗=∣a⃗∣2=42+(−1)2+32=16+1+9=26\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 4^2 + (-1)^2 + 3^2 = 16 + 1 + 9 = 26

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

Given that a⃗⋅c⃗=15\vec{a} \cdot \vec{c} = 15, we substitute these values into the dot product equation:

15=13(2(26)+35λ)15 = \frac{1}{3}\left(2(26) + 35\lambda\right)

45=52+35λ45 = 52 + 35\lambda

35λ=−7  ⟹  λ=−1535\lambda = -7 \implies \lambda = -\frac{1}{5}

Now, let d⃗=i^+j^−3k^\vec{d} = \hat{i} + \hat{j} - 3\hat{k}. We calculate the dot products a⃗⋅d⃗\vec{a} \cdot \vec{d} and b⃗⋅d⃗\vec{b} \cdot \vec{d}:

a⃗⋅d⃗=(4)(1)+(−1)(1)+(3)(−3)=4−1−9=−6\vec{a} \cdot \vec{d} = (4)(1) + (-1)(1) + (3)(-3) = 4 - 1 - 9 = -6

b⃗⋅d⃗=(10)(1)+(2)(1)+(−1)(−3)=10+2+3=15\vec{b} \cdot \vec{d} = (10)(1) + (2)(1) + (-1)(-3) = 10 + 2 + 3 = 15

Finally, we compute c⃗⋅d⃗\vec{c} \cdot \vec{d}:

c⃗⋅d⃗=13(2(a⃗⋅d⃗)+λ(b⃗⋅d⃗))\vec{c} \cdot \vec{d} = \frac{1}{3}\left(2\left(\vec{a} \cdot \vec{d}\right) + \lambda\left(\vec{b} \cdot \vec{d}\right)\right)

c⃗⋅d⃗=13(2(−6)+(−15)(15))\vec{c} \cdot \vec{d} = \frac{1}{3}\left(2(-6) + \left(-\frac{1}{5}\right)(15)\right)

c⃗⋅d⃗=13(−12−3)=−153=−5\vec{c} \cdot \vec{d} = \frac{1}{3}\left(-12 - 3\right) = \frac{-15}{3} = -5

Hence, c⃗⋅(i^+j^−3k^)=−5\vec{c} \cdot \left(\hat{i} + \hat{j} - 3\hat{k}\right) = -5, which corresponds to Option B.

Calculate Dot Product of Vector C with Given Unit Vector Combination | Mathematics PYQ Solution - JEE Challenger