Test Index

Vector Algebra

Section: Chemistry

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Question : 2 of 40

Marks: +1, -0

Let

\vec{a}, \vec{b}

be two vectors, and let P, Q and R be the points with position vectors

\vec{a}, \vec{b}

\vec{a} + \vec{b}

, respectively, with respect to the origin O. If

| \vec{a} + \vec{b} | = \sqrt{21}

| \vec{a} - \vec{b} | = 3

\vec{a}

(\vec{a} - \vec{b})

are perpendicular to each other, then the area of the triangle OPR is :

[JEE Adv 2026 P2]

$\sqrt{3}$
$\frac{\sqrt{3}}{2}$
$\frac{3\sqrt{3}}{2}$
$\frac{3}{2}$

Solution:

Concept:

Area of triangle

OPR

is

\frac{1}{2} | \vec{a} \times (\vec{a} + \vec{b}) | = \frac{1}{2} | \vec{a} \times \vec{b} |

.

We find

| \vec{a} \times \vec{b} |

using given magnitudes and dot product.

Explanation:

Given:

P=\vec{a}

,

R=\vec{a}+\vec{b}

,

O

is origin.

Area of

\triangle OPR = \frac{1}{2} |\overrightarrow{OP} \times \overrightarrow{OR}| = \frac{1}{2} |\vec{a} \times (\vec{a}+\vec{b})| = \frac{1}{2} |\vec{a} \times \vec{b}|

(since

\vec{a}\times\vec{a}=0

).

From

|\vec{a}+\vec{b}|=\sqrt{21}

, square:

|\vec{a}|^2+|\vec{b}|^2+2\vec{a}\cdot\vec{b}=21

...(1)

From

|\vec{a}-\vec{b}|=3

, square:

|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}=9

...(2)

Subtract (2) from (1):

4\vec{a}\cdot\vec{b}=12 \implies \vec{a}\cdot\vec{b}=3

.

\vec{a}

is perpendicular to

(\vec{a}-\vec{b})

:

\vec{a}\cdot(\vec{a}-\vec{b})=0 \implies |\vec{a}|^2 - \vec{a}\cdot\vec{b}=0 \implies |\vec{a}|^2=3

, so

|\vec{a}|=\sqrt{3}

.

Substitute into (2):

3+|\vec{b}|^2 -6 =9 \implies |\vec{b}|^2=12

.

Use

|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = (3)(12) - 3^2 = 36-9=27

.

Thus

|\vec{a}\times\vec{b}| = \sqrt{27}=3\sqrt{3}

.

Area =

\frac{1}{2} (3\sqrt{3}) = \frac{3\sqrt{3}}{2}

.

Answer:

\frac{3\sqrt{3}}{2}

(Option C).

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