The value of x for which the angle between th | vedclass

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(B) For the angle between vectors $a$ and $b$ to be acute, their dot product must be positive: $a \cdot b > 0$.$(-3i + xj + k) \cdot (xi + 2xj + k) >

Vedclass · Accepted Answer

$x < 0$

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(B) For the angle between vectors $a$ and $b$ to be acute, their dot product must be positive: $a \cdot b > 0$.$(-3i + xj + k) \cdot (xi + 2xj + k) > 0$$-3x + 2x^2 + 1 > 0$$2x^2 - 3x + 1 > 0$$(2x - 1)(x - 1) > 0$This inequality holds when $x 1$.For the angle between vector $b = xi + 2xj + k$ and the $x$-axis (represented by unit vector $i$) to be obtuse (between $\pi/2$ and $\pi$), the dot product $b \cdot i$ must be negative:$b \cdot i $(xi + 2xj + k) \cdot i $x Combining the conditions $x 1$ and $x < 0$, we get $x < 0$.

The value of $x$ for which the angle between the vectors $a = -3i + xj + k$ and $b = xi + 2xj + k$ is acute and the angle between $b$ and the $x$-axis lies between $\pi/2$ and $\pi$ satisfies:

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