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JEE MATHEMATICS MCQ: ADDITION OF VECTORS

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Let the position vectors of the points A,B,CA, B, C and DD be 5i^+5j^+2λk^,i^+2j^+3k^,−2i^+λj^+4k^5 \hat{i}+5 \hat{j} +2 \lambda \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k},-2 \hat{i}+\lambda \hat{j}+4 \hat{k} and −i^+5j^+6k^-\hat{i}+5 \hat{j}+6 \hat{k}. Let the set S={λ∈RS=\{\lambda \in \mathbb{R} : the points AA, B,CB, C and DD are coplanar}. Then ∑λ∈S(λ+2)2\sum\limits_{{\lambda \in S}}(\lambda+2)^{2} is equal to
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Correct option: D
41
The correct answer is D (41) because the sum of (λ+2)2(\lambda+2)^2 for the values of λ\lambda that make the points coplanar is 41.

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Let a=i^−j^,b=j^−k^\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}} and c=k^−i^\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}. If d\mathbf{d} is aa unit vector such that a⋅d=0=(bc\mathbf{a} \cdot \mathbf{d}=0=(b \mathbf{c} ) , then d\mathbf{d} is equal to
A vector a⃗{\mathbf{\vec a}} has components 2  p2\;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 a⃗{\mathbf{\vec a}} has components p+1p + 1 and 1, then If a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} are three mutually perpendicular unit vetors and d⃗\vec{d} is a unit vector which makes equal angles with a→,b→\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}} ānd c→\overrightarrow{\mathrm{c}} then the value of ∣a→+b→+c→+d→∣2|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|^2 The non-zero vectors are a⃗,b⃗\vec{a}, \vec{b} and c⃗\vec{c} are related by a⃗=8b⃗\vec{a}=8 \vec{b} and c⃗=−7b⃗\vec{c}=-7 \vec{b}. Then the angle between a⃗\vec{a} and c⃗\vec{c} is If the vectors 2i+j−k,−i+2j+λk2 \mathbf{i}+\mathbf{j}-\mathbf{k},-\mathbf{i}+2 \mathbf{j}+\lambda \mathbf{k} and −5i+2j−k-5 \mathbf{i}+2 \mathbf{j}-\mathbf{k} ar coplanar, then the value of λ\lambda is equal The medium ADAD of the triangle ABCABC is bisected at E,  BEE,\;BE meets ACAC in FF, then AF:AC=AF:AC =
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