If a=2i+j-k ,b= -i+2j-4k and c= i+j+k then find (a×b) . ( b×c )
Answers
Answer:
Step-by-step explanation:
(a×b)×(b×c)
=(2i+j-k × -i+2j-4k)×(-i+2j-4k × i+j+k)
=(-2i+2j+4k) × (-i+2j-4k)
=+2i+4j-16k
Answer:First, we need to make sure that the given vectors form a triangle. This will happen if the sum
of two of them is equal to the third vector, or if the sum of all three vectors is equal to the zero
vector.
In this case, A~ + B~ + C~ = ~0.
This triangle would be a right triangle if the angle between two vectors is equal to π/2, or
equivalently, if their dot product is equal to zero.
In this case, A~ · B~ = (−1)(−1) + (1)(−1) + (0)(−2) = 0. Therefore, the vectors A~ and B~ are
perpendicular and the triangle formed by A~, B~ , and C~ is a right triangle.
2. (25 pts) What is the value of the triple product A~ · (A~ × B~ )? Why?
Solution:
The value of A~ · (A~ × B~ ) = 0 because the result of the cross product A~ × B~ is a vector
perpendicular to both A~ and B~ , and since the dot product of two perpendicular vectors is zero,
the dot product of A~ and A~ × B~ is equal to zero.
3. (25 pts) Find the shortest distance from P(6, −4) to the line y = 2x − 3.
Solution:
If you recall, we solved a similar problem in homework 2 (problem 6). We found then that the
distance d between a point and a line in R
2
can be computed using d =
||AB~ ×AP~ ||
||AB~ || , where AB~
is a vector between two points A and B on the line, and AP~ is the vector from point A to the
point P we are interested in finding the distance to from the line.
Let A be the point on the line when x = 0 and B the point on the line when x = 1. So,
A(0, −3) and B(1, −1). Now, AB~ = h1, 2, 0i and AP~ = h6, −1, 0i. Computing the cross product
we obtain:
AB~ × AP~ =
ˆi ˆj
ˆk
1 2 0
6 −1 0
= 0ˆi + 0ˆj + (−1 − 12)ˆk = h0, 0, −13i.
Then, ||AB~ × AP~ || = 13 and ||AB~ || =
√
5. Thus, finally we can say that d = √
13
5
.
4. (25 pts) Find the area of a triangle with vertices at A(3, −1, 2), B(1, −1, −3), and C(4, −3, 1)
Solution:
We first define vectors to represent two sides of the triangle. So, AB~ = h−2, 0, −5i, and
AC~ = h1, −2, −1i. The area of the triangle is equal to a =
||AB~ ×AC~ ||
2
. So
AB~ × AC~ =
ˆi ˆj
ˆk
−2 0 −5
1 −2 −1
= −10ˆi − 7ˆj + 4ˆk = h−10, −7, 4i.
Then, ||AB~ × AC~ || =
p
(−10)2 + (−7)2 + (4)2 =
√
165, and a =
√
165
2
.
Bonus (10 pts): Simplify A~ · (2A~ + B~ ) × C~
Solution:
Using the properties of the dot and cross products, we have:
A~ ·(2A~+B~ )×C~ = A~ ·(2A~×C~ +B~ ×C~ ) = A~ ·(2A~×C~ )+A~ ·(B~ ×C~ ) = A~ ·(2(A~×C~ ))+A~ ·(B~ ×C~ ) =
2A~ · (A~ × C~ ) + A~ · (B~ × C~ ).
However, A~ · (A~ × C~ ) = 0 (see problem 2, above). Thus,
A~ · (2A~ + B~ ) × C~ = A~ · (B~ × C~ ).
Step-by-step explanation: