The vectors from the origin O to the points A and B isA - 3- - 6- - 2k- and B - 2- - - - 2k- respectively- The area of triangle OAB be

The vectors from origin to the points A and B are $\to a = 2^i - 3^j + 2^k a n d \to b = 2^i + 3^j +^k$ respectively,then the area of $△ O A B$ is equal to

(a) 340 (b) $\sqrt{25}$ (c) $\sqrt{229}$ (d) $\frac{1}{2} \sqrt{229}$

Q. Let

A, B, C

be points with position vectors

2^i -^j +^k,^i + 2^j +^k

and

3^i +^j + 2^k

respectively. Find the shortest distance between point

B

and plane

O A C

Q. If the position vector of the vertices of a triangle are

^i -^j + 2^k; 2^i +^j +^k

3^i -^j + 2^k

, then find the area of the triangle.

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The vectors from the origin O to the points A and B is→A=3^i−6^j+2^k and →B=2^i+^j−2^k respectively. The area of triangle OAB be

The correct option is A 52√17 sq. unitsArea of triangle =12|(→A×→B)| =12|(3^i−6^j+2^k)×(2^i+^j−2^k)| =12×|10^i+10^j+15^k| =52×(√17) sq. units

The vectors from the origin $O$ to the points $A$ and $B$ is $\to A = 3^i - 6^j + 2^k$ and $\to B = 2^i +^j - 2^k$ respectively. The area of triangle $O A B$ be

The correct option is A $\frac{5}{2} \sqrt{17}$ sq. units
Area of triangle $= \frac{1}{2} | (\to A \times \to B) |$
$= \frac{1}{2} | (3^i - 6^j + 2^k) \times (2^i +^j - 2^k) |$
$= \frac{1}{2} \times | 10^i + 10^j + 15^k |$
$= \frac{5}{2} \times (\sqrt{17})$ sq. units