Mathematics · 2019
JEE Main · 11 January 2019, Shift 1 · Q64
Let A = (0, 2 q, r; p, q, - r; p, - q, r). If AA^T = I_3, then | p | is:
Let $\displaystyle \mathrm{A}=\left(\begin{array}{ccc}0 & 2 \mathrm{q} & \mathrm{r} \\ \mathrm{p} & \mathrm{q} & -\mathrm{r} \\ \mathrm{p} & -\mathrm{q} & \mathrm{r}\end{array}\right)$. If $\displaystyle \mathrm{AA}^{\mathrm{T}}=\mathrm{I}_3$, then
$\displaystyle |\mathrm{p}|$ is :
Official answer
(1)
$\displaystyle \frac{1}{\sqrt{2}}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.