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:22 May Shift 2If a⃗\vec{a}a, b⃗\vec{b}b, c⃗\vec{c}c are unit vectors such that a⃗⋅b⃗=a⃗⋅c⃗=0\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0a⋅b=a⋅c=0, and the angle between b⃗\vec{b}b and c⃗\vec{c}c is π6\frac{\pi}{6}6π, then22 May Shift 1Let a⃗=i^+j^+k^\vec{a} = \hat{i} + \hat{j} + \hat{k}a=i^+j^+k^ and b⃗=i^+2j^+3k^\vec{b} = \hat{i} + \hat{2j} + 3\hat{k}b=i^+2j^+3k^ then a unit vector perpendicular to both vectors (a⃗+b⃗)(\vec{a} + \vec{b})(a+b) and (a⃗−b⃗)(\vec{a} - \vec{b})(a−b) is equal to15 May Shift 2If a⃗,b⃗\vec{a}, \vec{b}a,b and c⃗\vec{c}c are three unit vectors such that a⃗+2b⃗−3c⃗=0⃗\vec{a} + 2\vec{b} - 3\vec{c} = \vec{0}a+2b−3c=0, then the value of 2a⃗.b⃗−6b⃗.c⃗−3c⃗.a⃗2\vec{a}.\vec{b} - 6\vec{b}.\vec{c} - 3\vec{c}.\vec{a}2a.b−6b.c−3c.a is