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JEE Mathematics 2026 Question with Solution

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be vectors such that a⃗×b⃗=2(a⃗×c⃗)\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}). ∣a⃗∣=1,∣b⃗∣=4,∣c⃗∣=2|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2, angle between b⃗,c⃗\vec{b}, \vec{c} is 60∘60^\circ. Find ∣b⃗−2c⃗∣|\vec{b} - 2\vec{c}|.

  • A

    44

  • B

    22

  • C

    00

  • D

    11

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: a⃗×b⃗=2(a⃗×c⃗)\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}), ∣a⃗∣=1|\vec{a}| = 1, ∣b⃗∣=4|\vec{b}| = 4, ∣c⃗∣=2|\vec{c}| = 2, and the angle between b⃗\vec{b} and c⃗\vec{c} is 60∘60^\circ.

Find: ∣b⃗−2c⃗∣|\vec{b} - 2\vec{c}|.

From the given relation,

a⃗×b⃗−2a⃗×c⃗=0\vec{a} \times \vec{b} - 2\vec{a} \times \vec{c} = 0

so,

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

This implies b⃗−2c⃗\vec{b} - 2\vec{c} is parallel to a⃗\vec{a}, or

b⃗−2c⃗=λa⃗\vec{b} - 2\vec{c} = \lambda \vec{a}

Squaring both sides,

∣b⃗−2c⃗∣2=λ2∣a⃗∣2=λ2|\vec{b} - 2\vec{c}|^2 = \lambda^2 |\vec{a}|^2 = \lambda^2

Now expand the left-hand side:

∣b⃗−2c⃗∣2=∣b⃗∣2+4∣c⃗∣2−4b⃗⋅c⃗|\vec{b} - 2\vec{c}|^2 = |\vec{b}|^2 + 4|\vec{c}|^2 - 4\vec{b} \cdot \vec{c}

Also,

b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡60∘=4⋅2⋅12=4\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 60^\circ = 4 \cdot 2 \cdot \frac{1}{2} = 4

Therefore,

∣b⃗−2c⃗∣2=16+4(4)−4(4)=16|\vec{b} - 2\vec{c}|^2 = 16 + 4(4) - 4(4) = 16

Hence,

∣b⃗−2c⃗∣=4|\vec{b} - 2\vec{c}| = 4

Therefore, the correct option is A.

Using the vector modulus identity

Given: a⃗×b⃗=2(a⃗×c⃗)\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}).

Find: the modulus of b⃗−2c⃗\vec{b} - 2\vec{c}.

Use linearity of the cross product:

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

Hence, b⃗−2c⃗\vec{b} - 2\vec{c} is along a⃗\vec{a}. To find its magnitude, use the identity

∣u⃗−v⃗∣2=∣u⃗∣2+∣v⃗∣2−2u⃗⋅v⃗|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u} \cdot \vec{v}

Take u⃗=b⃗\vec{u} = \vec{b} and v⃗=2c⃗\vec{v} = 2\vec{c}. Then

∣b⃗−2c⃗∣2=∣b⃗∣2+∣2c⃗∣2−2b⃗⋅(2c⃗)|\vec{b} - 2\vec{c}|^2 = |\vec{b}|^2 + |2\vec{c}|^2 - 2\vec{b} \cdot (2\vec{c})

Now,

∣b⃗∣2=42=16,∣2c⃗∣2=(2∣c⃗∣)2=(2⋅2)2=16|\vec{b}|^2 = 4^2 = 16, \qquad |2\vec{c}|^2 = (2|\vec{c}|)^2 = (2 \cdot 2)^2 = 16

and

2b⃗⋅(2c⃗)=4(b⃗⋅c⃗)2\vec{b} \cdot (2\vec{c}) = 4(\vec{b} \cdot \vec{c})

Further,

b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡60∘=4⋅2⋅12=4\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 60^\circ = 4 \cdot 2 \cdot \frac{1}{2} = 4

So,

∣b⃗−2c⃗∣2=16+16−16=16|\vec{b} - 2\vec{c}|^2 = 16 + 16 - 16 = 16

Thus,

∣b⃗−2c⃗∣=4|\vec{b} - 2\vec{c}| = 4

So the required value is 44.

Common mistakes

  • Assuming a⃗×(b⃗−2c⃗)=0\vec{a} \times (\vec{b} - 2\vec{c}) = 0 forces b⃗−2c⃗=0\vec{b} - 2\vec{c} = 0. This is wrong because a zero cross product only implies the vectors are parallel or one of them is the zero vector. Instead, conclude that b⃗−2c⃗\vec{b} - 2\vec{c} is parallel to a⃗\vec{a} and then compute its magnitude.

  • Using the modulus formula incorrectly as ∣b⃗−2c⃗∣2=∣b⃗∣2+4∣c⃗∣2−2b⃗⋅c⃗|\vec{b} - 2\vec{c}|^2 = |\vec{b}|^2 + 4|\vec{c}|^2 - 2\vec{b} \cdot \vec{c}. This misses the factor coming from 2c⃗2\vec{c}. Instead, write the middle term carefully as −2b⃗⋅(2c⃗)=−4b⃗⋅c⃗-2\vec{b} \cdot (2\vec{c}) = -4\vec{b} \cdot \vec{c}.

  • Calculating b⃗⋅c⃗\vec{b} \cdot \vec{c} incorrectly by forgetting cos⁡60∘=12\cos 60^\circ = \frac{1}{2}. This gives the wrong final value. Instead, use b⃗⋅c⃗=∣b⃗∣∣c⃗∣cos⁡60∘=4⋅2⋅12=4\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos 60^\circ = 4 \cdot 2 \cdot \frac{1}{2} = 4.

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