- A$7$
- B$\sqrt{38}$
- C$\sqrt{155}$
- D$\text{none of these}$
Solution:
The given point (2, 3, 5) and (5, 9, 7) are two diagonally opposite vertices of the parallelopiped as all of theire coordinates are different.
$\therefore$ Edges of the paralleloppiped
= |2 - 5|, |3 - 9| and |5 - 7|
=3, 6 and 2.
Now,
Length of the diagonal of the parallelopiped
$=\sqrt{3^2+6^2+2^2}$
$=\sqrt{9+36+4}$
$=\sqrt{49}$
$=7$
Hence, length of the diagonal of the parallelepiped formed by the planes
Parallel to coordinate planes and drawn through point (2, 3, 5)and (5, 9, 7) is 7 units.
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$(P)$ If $A \neq I_{2},$ then $|A|=-1$
$(\mathrm{Q})$ If $|\mathrm{A}|=1,$ then $\operatorname{tr}(\mathrm{A})=2$
where $I_{2}$ denotes $2 \times 2$ identity matrix and $\operatorname{tr}(A)$ denotes the sum of the diagonal entries of $A$ Then