If the components of \veca=\alpha \hati+\beta \hatj+\gamma \hatk along and perpendicular to \vecb=3 \hati+\hatj-\hatk respectively, are \frac1611(3 \hati+\hatj-\hatk) and \frac111(-4 \hati-5 \hatj-17 \hatk) , then \alpha^2+\beta^2+\gamma^2 is equal to :

Vector Algebra PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q33 with Solution

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Correct answer: C

The vector $\vec a$ is resolved into:

So,

\vec a = \vec a_{\parallel} + \vec a_{\perp}

with

\vec a_{\parallel} = \frac{16}{11}(3\hat i + \hat j - \hat k)

and

\vec a_{\perp} = \frac{1}{11}(-4\hat i - 5\hat j - 17\hat k)

First compute the parallel component:

\vec a_{\parallel} = \left(\frac{48}{11}\right)\hat i + \left(\frac{16}{11}\right)\hat j - \left(\frac{16}{11}\right)\hat k

Now add the perpendicular component:

\vec a = \frac{1}{11}(48\hat i + 16\hat j - 16\hat k) + \frac{1}{11}(-4\hat i - 5\hat j - 17\hat k)

\vec a = \frac{1}{11}(44\hat i + 11\hat j - 33\hat k)

\vec a = 4\hat i + \hat j - 3\hat k

Hence,

\alpha = 4,\quad \beta = 1,\quad \gamma = -3

\alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2

= 16 + 1 + 9 = 26

The value is

26

So the correct option is C.

Stored correct answer: C

Our derived answer also gives C. So they agree.