Question

If the vectors (i +i+k) and 3î form two sides of
a triangle, then 3rd side of the triangle can be :
(1) /18 unit
(2) 6 unit
3
(3)
V2
(4) both 1 & 2
unit​

Answer 1

Answer:

In vector land, the triangle area is half the magnitude of the cross product,

(1,1,1)×(3,0,0)=(0,−(1(0)−3(1)),1(0)−3(1))=(0,3,−3)(1,1,1)×(3,0,0)=(0,−(1(0)−3(1)),1(0)−3(1))=(0,3,−3)

so

A=1202+32+32−−−−−−−−−−√=1218−−√=322–√A=1202+32+32=1218=322

That’s none of the above. 32–√32 is the area of the associated parallelogram.

Another way to see it is Archimedes’ Theorem. A triangle with squared sides A,B,CA,B,C has area SS satisfying

16S2=4AB−(A+B−C)216S2=4AB−(A+B−C)2

We have a triangle with vertices (0,0,0),(1,1,1),(3,0,0)(0,0,0),(1,1,1),(3,0,0)

A=12+12+12=3A=12+12+12=3

B=32+02+02=9B=32+02+02=9

C=(1−3)2+12+12=6C=(1−3)2+12+12=6

We see it’s a right triangle, satisfying the Pythagorean Theorem, here B=A+CB=A+C

We don’t care. The formula works for any triangle. But we can rewrite it to make BB the odd side out to make the arithmetic a tiny bit simpler.

16S2=4AC−(A+C−B)216S2=4AC−(A+C−B)2

16S2=4(3)(6)−(3+6−9)2=233216S2=4(3)(6)−(3+6−9)2=2332

4S=62–√4S=62

S=322–√✓S=322✓

The actual answer is 3 / sqrt(2).

assuming i,j,k are unit vectors in euclidean 3-D space and not quaternions:-;

length of 3i clearly is 3

length of i+j+k is sqrt(3)

length of the side from 3i to i+j+k = length (-2i+j+k) = sqrt(6).

Now apply Herons formula, and you will get something like sqrt ( 4 * 3 * 6 / 16) or 3 / sqrt(2).

Or you can just take the cross products of the vectors (1,1,1) and (3,0,0) which will be (0,3,-3). That has a length of sqrt(18). The area of the triangle is sqrt(18)/2.

Explanation:

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