JEEOlympiad

Let a = (2, 1, -2) and b = (1, 3, 2). Over all unit vectors c that…

Question

Let a = (2, 1, -2) and b = (1, 3, 2). Over all unit vectors c that are perpendicular to w = (1, 1, 1), the scalar triple product [a b c] attains a maximum value M. Then M^2 = m/n in lowest terms with m, n positive integers. Find m + n.

✓ Verified answer: 329checked by our engine — not a guess

Step-by-step solution

Note [a b c] = c·(a × b) = c·v where v = a × b.

We maximize c·v over unit vectors c with c·w = 0; the maximum is the length of the projection of v onto the plane perpendicular to w, namely M = sqrt(|v|^2 - (v·w)^2/|w|^2), so M^2 = |v|^2 - (v·w)^2/|w|^2.

v = a × b = (1·2 - (-2)·3, (-2)·1 - 2·2, 2·3 - 1·1) = (8, -6, 5).
|v|^2 = 64 + 36 + 25 = 125; v·w = 8 - 6 + 5 = 7; |w|^2 = 3.
M^2 = 125 - 49/3 = (375 - 49)/3 = 326/3.
Then m + n = 326 + 3 = 329.

Final answer329

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