The least positive integral value of α, for which the angle between the vectors α î-2 ĵ+2 k̂ and α î+2 α ĵ-2 k̂ is acute, is ___________.

Solution

Maths sir · 4 steps

Acute angle means a positive dot product, so set it above zero and solve the quadratic in α.

1. Condition for acute. The angle is acute when cos θ > 0. The magnitudes are always positive, so the dot product itself must be positive.
2. Find the dot product. Multiply matching components and add: (α)(α) + (-2)(2α) + (2)(-2) = α² - 4α - 4.

   α² - 4α - 4 > 0 ⇒ α < 2 - 2√2 or α > 2 + 2√2

3. Locate the roots. The roots are 2 ± 2√2, about -0.83 and 4.83. Every positive α that works must exceed 4.83. Check: α = 4 gives 16 - 16 - 4 = -4, which fails, and α = 5 gives 25 - 20 - 4 = 1, which works.
4. So. At α = 5 the vectors (5, -2, 2) and (5, 10, -2) are not parallel, so the angle is truly acute. The least positive integer is 5.

The common slip: 1. Taking the k̂ product (2)(-2) as +4 gives (α - 2)² > 0, which allows α = 1, not 5.

The idea to keep. Acute means dot product greater than 0, obtuse means less than 0. Turn the angle condition into a sign condition, then solve the inequality and test the integers on either side of the root.

Answer from the NTA final answer key. Solution by PrepWiser.