Question 1

SCQMEDIUM

Let a⃗,b⃗\vec{a}, \vec{b} be two vectors, and let P,QP, Q and RR be the points with position vectors a⃗,b⃗\vec{a}, \vec{b} and a⃗+b⃗\vec{a} + \vec{b}, respectively, with respect to the origin OO. If ∣a⃗+b⃗∣=21|\vec{a} + \vec{b}| = \sqrt{21}, ∣a⃗−b⃗∣=3|\vec{a} - \vec{b}| = 3, and a⃗\vec{a} and (a⃗−b⃗)(\vec{a} - \vec{b}) are perpendicular to each other, then the area of the triangle OPROPR is

(A)

3\sqrt{3}

(B)

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

(C)

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

(D)

32\frac{3}{2}

Detailed Solution

Given ∣a⃗+b⃗∣=21  ⟹  ∣a⃗+b⃗∣2=21  ⟹  ∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗=21|\vec{a} + \vec{b}| = \sqrt{21} \implies |\vec{a} + \vec{b}|^2 = 21 \implies |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 21 ---(i) Given ∣a⃗−b⃗∣=3  ⟹  ∣a⃗−b⃗∣2=9  ⟹  ∣a⃗∣2+∣b⃗∣2−2a⃗⋅b⃗=9|\vec{a} - \vec{b}| = 3 \implies |\vec{a} - \vec{b}|^2 = 9 \implies |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 9 ---(ii) Adding (i) and (ii): 2(∣a⃗∣2+∣b⃗∣2)=30  ⟹  ∣a⃗∣2+∣b⃗∣2=152(|\vec{a}|^2 + |\vec{b}|^2) = 30 \implies |\vec{a}|^2 + |\vec{b}|^2 = 15. Subtracting (ii) from (i): 4a⃗⋅b⃗=12  ⟹  a⃗⋅b⃗=34\vec{a} \cdot \vec{b} = 12 \implies \vec{a} \cdot \vec{b} = 3. Since a⃗\vec{a} is perpendicular to (a⃗−b⃗)(\vec{a} - \vec{b}), we have a⃗⋅(a⃗−b⃗)=0  ⟹  ∣a⃗∣2−a⃗⋅b⃗=0  ⟹  ∣a⃗∣2=3\vec{a} \cdot (\vec{a} - \vec{b}) = 0 \implies |\vec{a}|^2 - \vec{a} \cdot \vec{b} = 0 \implies |\vec{a}|^2 = 3. Then ∣b⃗∣2=15−3=12|\vec{b}|^2 = 15 - 3 = 12. Area of triangle OPR=12∣OP⃗×OR⃗∣=12∣a⃗×(a⃗+b⃗)∣=12∣a⃗×a⃗+a⃗×b⃗∣=12∣a⃗×b⃗∣OPR = \frac{1}{2} |\vec{OP} \times \vec{OR}| = \frac{1}{2} |\vec{a} \times (\vec{a} + \vec{b})| = \frac{1}{2} |\vec{a} \times \vec{a} + \vec{a} \times \vec{b}| = \frac{1}{2} |\vec{a} \times \vec{b}|. Now, ∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2=(3)(12)−(3)2=36−9=27|\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. So, ∣a⃗×b⃗∣=27=33|\vec{a} \times \vec{b}| = \sqrt{27} = 3\sqrt{3}. Area of △OPR=332\triangle OPR = \frac{3\sqrt{3}}{2}.

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