CUET MathematicsAlgebra > Mediumcore16i^+26j^+16k^\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{\mathrm{j}}+\frac{1}{\sqrt{6}} \hat{\mathrm{k}}61i^+62j^+61k^−16i^+16j^−16k^-\frac{1}{\sqrt{6}} \hat{\mathrm{i}}+\frac{1}{\sqrt{6}} \hat{\mathrm{j}}-\frac{1}{\sqrt{6}} \hat{\mathrm{k}}−61i^+61j^−61k^−16i^+26j^+26k^-\frac{1}{\sqrt{6}} \hat{\mathrm{i}}+\frac{2}{\sqrt{6}} \hat{\mathrm{j}}+\frac{2}{\sqrt{6}} \hat{\mathrm{k}}−61i^+62j^+62k^−16i^+26j^−16k^-\frac{1}{\sqrt{6}} \hat{\mathrm{i}}+\frac{2}{\sqrt{6}} \hat{\mathrm{j}}-\frac{1}{\sqrt{6}} \hat{\mathrm{k}}−61i^+62j^−61k^✅ Correct Option: 4Related questions:3 June Shift 2Let a⃗=i^+2j^+3k^,b⃗=−i^+2j^+k^,c⃗=3i^+j^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = -\hat{i} + 2\hat{j} + \hat{k}, \vec{c} = 3\hat{i} + \hat{j}a=i^+2j^+3k^,b=−i^+2j^+k^,c=3i^+j^ be three vectors. If (a⃗+λb⃗)(\vec{a} + \lambda\vec{b})(a+λb) is perpendicular to c⃗\vec{c}c, then the value of λ\lambdaλ is21 May Shift 2The area of triangle with vertices P, Q, R is given by (where AB⃗\vec{AB}AB = position vector of point B – position vector of point A)3 June Shift 2If a⃗\vec{a}a and b⃗\vec{b}b are two vectors such that ∣a⃗∣=10,∣b⃗∣=2|\vec{a}| = 10, |\vec{b}| = 2∣a∣=10,∣b∣=2 and a⃗⋅b⃗=12\vec{a} \cdot \vec{b} = 12a⋅b=12, then ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|∣a×b∣ is equal to