JEE Main 2023 — Mathematics Vectors & 3D Geometry

2023mcqhard

For any vector $\vec{a}={a}_{1}\hat{i}+{a}_{2}\hat{j}+{a}_{3}\hat{k}$ , with $10|{a}_{i}|<1,i=1,2,3$ , consider the following statements:

$(A)$ : $\mathrm{max}{|{a}_{1}|,|{a}_{2}|,|{a}_{3}|}\leq |\vec{a}|$

$(B)$ : $|\vec{a}|\leq 3\mathrm{max}{|{a}_{1}|,|{a}_{2}|,|{a}_{3}|}$

Official previous-year question

Held on 11 Apr 2023 · Verified 6 Jul 2026.

Options

A
Only $(B)$ is true
B
Only $(A)$ is true
C
Both $(A)$ and $(B)$ are true
D
Neither $(A)$ nor $(B)$ is true

Did you get this right?

Solution

Given,

$\vec{a}={a}_{1}\hat{i}+{a}_{2}\hat{j}+{a}_{3}\hat{k}$ , with $10|{a}_{i}|<1,i=1,2,3$ ,

Let us assume that $|{a}_{1}|\leq |{a}_{2}|\leq |{a}_{3}|$

Now we know that,

${|\vec{a}|}^{2}={|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

And ${|{\vec{a}}_{3}|}^{2}\leq {|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

So, combining both above equations we get,

$\Rightarrow {|\vec{a}|}^{2}\geq {|{\vec{a}}_{3}|}^{2}$

$\Rightarrow |\vec{a}|\geq |{\vec{a}}_{3}|=\text{max}{|{\vec{a}}_{1}|,|{\vec{a}}_{2}||{\vec{a}}_{3}|}$

Hence, $(A)$ is true

Now again solving,

${|\vec{a}|}^{2}={|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

And ${|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}\leq {|{\vec{a}}_{3}|}^{2}+{|{\vec{a}}_{3}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

$\Rightarrow {|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}\leq 3{|{\vec{a}}_{3}|}^{2}$

$\Rightarrow {|\vec{a}|}^{2}\leq 3{|{\vec{a}}_{3}|}^{2}$

$\Rightarrow |\vec{a}|\leq 3|{\vec{a}}_{3}|=\text{max}{|{\vec{a}}_{1}|,|{\vec{a}}_{2}||{\vec{a}}_{3}|}$

Hence, $(B)$ is also true.

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Solution

Given,

$\vec{a}={a}_{1}\hat{i}+{a}_{2}\hat{j}+{a}_{3}\hat{k}$ , with $10|{a}_{i}|<1,i=1,2,3$ ,

Let us assume that $|{a}_{1}|\leq |{a}_{2}|\leq |{a}_{3}|$

Now we know that,

${|\vec{a}|}^{2}={|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

And ${|{\vec{a}}_{3}|}^{2}\leq {|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

So, combining both above equations we get,

$\Rightarrow {|\vec{a}|}^{2}\geq {|{\vec{a}}_{3}|}^{2}$

$\Rightarrow |\vec{a}|\geq |{\vec{a}}_{3}|=\text{max}{|{\vec{a}}_{1}|,|{\vec{a}}_{2}||{\vec{a}}_{3}|}$

Hence, $(A)$ is true

Now again solving,

${|\vec{a}|}^{2}={|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

And ${|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}\leq {|{\vec{a}}_{3}|}^{2}+{|{\vec{a}}_{3}|}^{2}+{|{\vec{a}}_{3}|}^{2}$

$\Rightarrow {|{\vec{a}}_{1}|}^{2}+{|{\vec{a}}_{2}|}^{2}+{|{\vec{a}}_{3}|}^{2}\leq 3{|{\vec{a}}_{3}|}^{2}$

$\Rightarrow {|\vec{a}|}^{2}\leq 3{|{\vec{a}}_{3}|}^{2}$

$\Rightarrow |\vec{a}|\leq 3|{\vec{a}}_{3}|=\text{max}{|{\vec{a}}_{1}|,|{\vec{a}}_{2}||{\vec{a}}_{3}|}$

Hence, $(B)$ is also true.

JEE Main 2023 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

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JEE Main Vectors & 3D Geometry in other years

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JEE Main 2023 — Mathematics Vectors & 3D Geometry

Options

Did you get this right?

Solution

Your note

JEE Main Vectors & 3D Geometry in other years

More JEE Main Vectors & 3D Geometry questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

Keep practicing Vectors & 3D Geometry