Mathematics · Vector Algebra

JEE Main 2024 — 5 April, Shift 2 — Question 7

Consider three vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}. Let ∣a⃗∣=2,∣b⃗∣=3|\vec{a}|=2,|\vec{b}|=3 and a⃗=b⃗×c⃗\vec{a}=\vec{b} \times \vec{c}. If α∈[0,π3]\alpha \in\left[0, \frac{\pi}{3}\right] is the angle between the vectors b⃗\vec{b} and c⃗\vec{c}, then the minimum value of 27∣c→−a→∣227|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|^{2} is equal to :

  1. Option A:

    110

  2. Option B:

    105

  3. Option C:

    124

    Correct
  4. Option D:

    121

Answer: C

Step-by-step solution

∣c→−a→∣=∣c→∣2+∣a→∣2−2a‾⋅c‾|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{c}}|^{2}+|\overrightarrow{\mathrm{a}}|^{2}-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}

=∣c⃗∣2+4−0=|\vec{c}|^{2}+4-0

∵a→=b→×c→\because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} ∣a→∣=∣b→×c→∣|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}| 2=3∣c→∣sin⁡α2=3|\overrightarrow{\mathrm{c}}| \sin \alpha ∣c→∣=23cosec⁡α|\overrightarrow{\mathrm{c}}|=\frac{2}{3} \operatorname{cosec} \alpha

α∈[0,π3]\alpha \in\left[0, \frac{\pi}{3}\right] ∣c→∣min⁡=23×23|\overrightarrow{\mathrm{c}}|_{\min }=\frac{2}{3} \times \frac{2}{\sqrt{3}}

cosec⁡α∈[23,∞)\operatorname{cosec} \alpha \in\left[\frac{2}{\sqrt{3}}, \infty\right)

⇒27∣c→−a⃗∣min⁡2=27(1627+4)=124\Rightarrow 27|\overrightarrow{\mathrm{c}}-\vec{a}|_{\min }^{2}=27\left(\frac{16}{27}+4\right)=124

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Vector Algebra
Topic
Vector or Cross Product of Two Vectors