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:13 May Shift 2A vector of magnitude 8 units in the direction perpendicular to both the vectors i^+j^+k^\hat{i} + \hat{j} + \hat{k}i^+j^+k^ and 2i^+k^2\hat{i} + \hat{k}2i^+k^ is21 May Shift 1If a⃗\vec{a}a, b⃗\vec{b}b and 3a⃗−b⃗\sqrt{3}\vec{a} - \vec{b}3a−b are three unit vectors, then the angle between a⃗\vec{a}a and b⃗\vec{b}b is:21 May Shift 2If a⃗=2j^−k^\vec{a} = 2\hat{j} - \hat{k}a=2j^−k^, b⃗=2i^−3j^+k^\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}b=2i^−3j^+k^ and c⃗=−i^+k^\vec{c} = -\hat{i} + \hat{k}c=−i^+k^ are three vectors, then the area (in sq. units) of the parallelogram whose diagonals are (b⃗+c⃗)(\vec{b} + \vec{c})(b+c) and (a⃗+c⃗)(\vec{a} + \vec{c})(a+c) is