If the angle θ between the vectors a→=2x2i→+4xj→+k→ and b→=7i→−2j→+xk→ is such that 90∘<θ<180∘ then x lies in the interval.

90<θ<180, θ is obtuse a¯.b¯<02x2i¯+4xj¯+k.7i¯−2j¯+xk¯<014x2−8x+x<014x2−7x<07x(2x−1)<0x(2x−1)<0(x−0)x−12<0x∈0, 12

If the angle θ between the vectors a→=2x2i→+4xj→+k→ and b→=7i→−2j→+xk→ is such that 90∘<θ<180∘ then x lies in the interval.

90<θ<180, θ is obtuse a¯.b¯<02x2i¯+4xj¯+k.7i¯−2j¯+xk¯<014x2−8x+x<014x2−7x<07x(2x−1)<0x(2x−1)<0(x−0)x−12<0x∈0, 12

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If the angle $θ$ between the vectors $\vec{a} = 2 x^{2} \vec{i} + 4 x \vec{j} + \vec{k} $ and $\vec{b} = 7 \vec{i} - 2 \vec{j} + x \vec{k} $ is such that $90^{\circ} < θ < 180^{\circ}$ then x lies in the interval.

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$(\frac{1}{2}, \frac{3}{2})$

$(\frac{1}{2}, 1)$

$(1, \frac{3}{2})$

$(0, \frac{1}{2})$

answer is A.

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Detailed Solution

$90 < θ < 180, θ$ is obtuse $\bar{a} . \bar{b} < 0$

$(2 x^{2} \bar{i} + 4 x \bar{j} + k) . (7 \bar{i} - 2 \bar{j} + x \bar{k}) < 0$

$14 x^{2} - 8 x + x < 0$

$14 x^{2} - 7 x < 0$

$7 x (2 x - 1) < 0$

$x (2 x - 1) < 0$

$(x - 0) (x - \frac{1}{2}) < 0$

$x \in (0, \frac{1}{2})$

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