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JEE MATHEMATICS MCQ: ADDITION OF VECTORS

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If ∣x1∣>∣y1∣+∣z1∣,∣y2∣>∣x2∣+∣z2∣,∣z3∣>∣x3∣+∣y3∣\left|x_1\right|>\left|y_1\right|+\left|z_1\right|,\left|y_2\right|>\left|x_2\right|+\left|z_2\right|,\left|z_3\right|>\left|x_3\right|+\left|y_3\right|,
then x1i^+y1j^+z1k^,x2i^+y2j^+z2k^x_1 \hat{\mathbf{i}}+y_1 \hat{\mathbf{j}}+z_1 \hat{\mathbf{k}}, x_2 \hat{\mathbf{i}}+y_2 \hat{\mathbf{j}}+z_2 \hat{\mathbf{k}} and x3i^+y3j^+z3k^x_3 \hat{\mathbf{i}}+y_3 \hat{\mathbf{j}}+z_3 \hat{\mathbf{k}} are
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Correct option: D
non-coplanar
The vectors are non-coplanar because the given inequalities contradict the condition for coplanarity, which requires the determinant to be zero.

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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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