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JEE MATHEMATICS MCQ: ADDITION OF VECTORS

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A unit tangent vector at t=2t=2 on the curve x=t2+2,y=4t3βˆ’5,z=2t2βˆ’6tx=t^2+2, y=4 t^3-5, z=2 t^2-6 t is
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Correct option: D
None of these
Option D (None of these) is correct because the calculated unit tangent vector at t=2 does not match any of the given options.

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Let a=i^βˆ’j^,b=j^βˆ’k^\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}} and c=k^βˆ’i^\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}. If d\mathbf{d} is aa unit vector such that aβ‹…d=0=(bc\mathbf{a} \cdot \mathbf{d}=0=(b \mathbf{c} ) , then d\mathbf{d} is equal to
 A vector aβƒ—{\mathbf{\vec a}} has components 2β€…β€Šp2\;p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense, if this respect to new system aβƒ—{\mathbf{\vec a}} has components p+1p + 1 and 1, then If aβƒ—,bβƒ—\vec{a}, \vec{b} and cβƒ—\vec{c} are three mutually perpendicular unit vetors and dβƒ—\vec{d} is a unit vector which makes equal angles with aβ†’,bβ†’\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} ānd cβ†’\overrightarrow{\mathrm{c}} then the value of ∣aβ†’+bβ†’+cβ†’+dβ†’βˆ£2|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|^2 The non-zero vectors are aβƒ—,bβƒ—\vec{a}, \vec{b} and cβƒ—\vec{c} are related by aβƒ—=8bβƒ—\vec{a}=8 \vec{b} and cβƒ—=βˆ’7bβƒ—\vec{c}=-7 \vec{b}. Then the angle between aβƒ—\vec{a} and cβƒ—\vec{c} is If the vectors 2i+jβˆ’k,βˆ’i+2j+Ξ»k2 \mathbf{i}+\mathbf{j}-\mathbf{k},-\mathbf{i}+2 \mathbf{j}+\lambda \mathbf{k} and βˆ’5i+2jβˆ’k-5 \mathbf{i}+2 \mathbf{j}-\mathbf{k} ar coplanar, then the value of Ξ»\lambda is equal The medium ADAD of the triangle ABCABC is bisected at E,β€…β€ŠBEE,\;BE meets ACAC in FF, then AF:AC=AF:AC =
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