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JEE MATHEMATICS MCQ: ADDITION OF VECTORS

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If a⃗,b⃗,c⃗\mathbf{\vec a},\mathbf{\vec b},\mathbf{\vec c} are the position vectors of the vertices of an equilateral triangle, whose orthocenter is at the origin, then
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Correct option: A
a⃗+b⃗+c⃗=0⃗\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c} = \vec 0
The correct answer is A (a⃗+b⃗+c⃗=0⃗\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c} = \vec 0) because in an equilateral triangle, the centroid and orthocenter coincide. If the orthocenter is at the origin, the centroid is also at the origin, implying the sum of the position vectors is zero.

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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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