How to solve 2(a×b)+3(b×c)=0 for c?

Rearrange to (2a−3c)×b=0, so 2a−3c=λb. Then c=(2a−λb)/3. Use a·c=15 to find λ=1/5, then compute c.

What is c·(i+j−3k)?

c=2i−(4/5)j+(31/15)k. c·(i+j−3k)=2−4/5−31/5=2−7=−5.

a⃗=4î−ĵ+3k̂, b⃗=10î+2ĵ−k̂, 2(a×b)+3(b×c)=0, a·c=15 — find c·(î+ĵ−3k̂)

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Let $\vec{a}=4\hat{i}-\hat{j}+3\hat{k}$, $\vec{b}=10\hat{i}+2\hat{j}-\hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a}\times\vec{b})+3(\vec{b}\times\vec{c})=\vec{0}$. If $\vec{a}\cdot\vec{c}=15$, then $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})$ is equal to:

A) $-6$ B) $-5$ C) $-4$ D) $-3$

✅ Correct Answer

B) −5

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Solution

Simplify the vector equation

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

$\Rightarrow\quad 2(\vec{a}\times\vec{b})=-3(\vec{b}\times\vec{c})=3(\vec{c}\times\vec{b})$

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

This means $2\vec{a}-3\vec{c}$ is parallel to $\vec{b}$, so $2\vec{a}-3\vec{c}=\lambda\vec{b}$ for some scalar $\lambda$.

Find λ using a·c = 15

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

$\vec{a}\cdot\vec{c}=\dfrac{2|\vec{a}|^2-\lambda(\vec{a}\cdot\vec{b})}{3}=15$

$|\vec{a}|^2=16+1+9=26,\quad\vec{a}\cdot\vec{b}=40-2-3=35$

$\dfrac{2(26)-35\lambda}{3}=15\quad\Rightarrow\quad52-35\lambda=45\quad\Rightarrow\quad\lambda=\dfrac{7}{35}=\dfrac{1}{5}$

Find c⃗

$\vec{c}=\dfrac{2\vec{a}-\frac{1}{5}\vec{b}}{3}=\dfrac{2(4\hat{i}-\hat{j}+3\hat{k})-\frac{1}{5}(10\hat{i}+2\hat{j}-\hat{k})}{3}$

$=\dfrac{(8-2)\hat{i}+(-2-\frac{2}{5})\hat{j}+(6+\frac{1}{5})\hat{k}}{3}=\dfrac{6\hat{i}-\frac{12}{5}\hat{j}+\frac{31}{5}\hat{k}}{3}$

$=2\hat{i}-\dfrac{4}{5}\hat{j}+\dfrac{31}{15}\hat{k}$

Compute c⃗·(î+ĵ−3k̂)

$\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})=2+\left(-\dfrac{4}{5}\right)+(-3)\cdot\dfrac{31}{15}$

$=2-\dfrac{4}{5}-\dfrac{31}{5}=2-\dfrac{35}{5}=2-7=\boxed{-5}$

📘 Parallel Vector Condition from Cross Product

If $\vec{u}\times\vec{v}=\vec{0}$, then $\vec{u}\parallel\vec{v}$. Use this to convert cross product equations into scalar multiples. Then use dot product conditions to find the scalar. This converts a vector equation into two algebraic equations.

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Important Concepts

Cross Product Zero Implies Parallel$(2\vec{a}-3\vec{c})\times\vec{b}=\vec{0}$ means $2\vec{a}-3\vec{c}=\lambda\vec{b}$. This is the key step that eliminates the cross product and gives a linear relation between the vectors.

Computing |a|² and a·b$|\vec{a}|^2=4^2+(-1)^2+3^2=16+1+9=26$. $\vec{a}\cdot\vec{b}=4(10)+(-1)(2)+3(-1)=40-2-3=35$. These scalar values are needed to find $\lambda$.

Finding λ from Dot ProductTake dot product of $\vec{c}=(2\vec{a}-\lambda\vec{b})/3$ with $\vec{a}$: $\vec{a}\cdot\vec{c}=(2|\vec{a}|^2-\lambda\vec{a}\cdot\vec{b})/3=(52-35\lambda)/3=15$. Solving: $\lambda=1/5$.

Final Dot Product Calculation$\vec{c}=2\hat{i}-\frac{4}{5}\hat{j}+\frac{31}{15}\hat{k}$. Then $\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})=2-4/5-93/15=2-4/5-31/5=2-35/5=2-7=-5$.

FAQs

How does 2(a×b)+3(b×c)=0 simplify?

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$3(\vec{b}\times\vec{c})=-2(\vec{a}\times\vec{b})=2(\vec{b}\times\vec{a})$. So $3(\vec{b}\times\vec{c})-2(\vec{b}\times\vec{a})=\vec{b}\times(3\vec{c}-2\vec{a})=\vec{0}$, meaning $3\vec{c}-2\vec{a}\parallel\vec{b}$.

How to dot a with c when c=(2a−λb)/3?

⌄

$\vec{a}\cdot\vec{c}=\vec{a}\cdot\frac{2\vec{a}-\lambda\vec{b}}{3}=\frac{2|\vec{a}|^2-\lambda(\vec{a}\cdot\vec{b})}{3}=\frac{2(26)-35\lambda}{3}=15$.

What is c in component form?

⌄

$\vec{c}=2\hat{i}-\frac{4}{5}\hat{j}+\frac{31}{15}\hat{k}$.

Verify: does a·c=15?

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$\vec{a}\cdot\vec{c}=4(2)+(-1)(-4/5)+3(31/15)=8+4/5+31/5=8+35/5=8+7=15$ ✓

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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