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Questions >> TN XII Math >> Vector Algebra
Step 1
A(3, –1, 2), B(1, –1, –3), C(4, –3, 1)
Here vector area of triangle =
H(АВ х АС)
AB = (1 – 3) i + (-1+1)
+(-3 – 2) k
-2 i - 5 k
AC = (4 – 3) +(-3 + 1)
+ (1 – 2) k
i - 2 i - k
Step 2
AB x AC =
-2
-5
1
-2 -1
= (0 – 10) i
7-(2 + 5) + (4 – 0) k
= -10 i +7 i: 4 k
Vector area
A ABC = }(-10 i +7j +4 k )
Area =V100 + 49 + 16
V165
sq.
units
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