Mathematics · Vector Algebra

JEE Advanced 2020 — Paper 1 — Question 43

In a triangle PQRP Q R, let a⃗=QR→,b⃗=RP→\vec{a}=\overrightarrow{Q R}, \vec{b}=\overrightarrow{R P} and c⃗=PQ→\vec{c}=\overrightarrow{P Q}. If

∣a→∣=3,∣ b→∣=4 and a→⋅(c→−b→)c→⋅(a→−b→)=∣a→∣∣a→∣+∣b→∣,|\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{~b}}|=4 \text { and } \frac{\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}})}{\overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})}=\frac{|\overrightarrow{\mathrm{a}}|}{|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{b}}|}, then the value of ∣a⃗×b⃗∣2|\vec{a} \times \vec{b}|^{2} is ____\_\_\_\_

Answer: 108

Numerical answer — enter this value.

Step-by-step solution

Equation reduces to (4a⃗+3b⃗)⋅c⃗=7a⃗⋅b⃗(4 \vec{a}+3 \vec{b}) \cdot \vec{c}=7 \vec{a} \cdot \vec{b}

But a⃗+b⃗+c⃗=0\vec{a}+\vec{b}+\vec{c}=0

So, (4a⃗+3b⃗)⋅(a⃗+b⃗)=−7a⃗⋅b⃗(4 \vec{a}+3 \vec{b}) \cdot(\vec{a}+\vec{b})=-7 \vec{a} \cdot \vec{b}

a⃗⋅b⃗=−6\vec{a} \cdot \vec{b}=-6

∣a⃗×b⃗∣2=108|\vec{a} \times \vec{b}|^{2}=108

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Mathematics
Chapter
Vector Algebra
Topic
Vector or Cross Product of Two Vectors