if A=2i+21+3k and B=3i -21-4k find the dot product
Answers
Answer:
Explanation:
A⃗ =−2i^+3j^ and
B⃗ =3i^+2j^
Dot product of two vector is
A⃗ ⋅B⃗ =(−2i^+3j^)⋅(B⃗ =3i^+2j^)
A⃗ ⋅B⃗ =(−2)(3)+(3)(2)=0
The dot product of two vector is zero which implies that they are perpendicular to each other. This can be obtained as fallows
A⃗ ⋅B⃗ =0
ABcosθ=0
Where A and B are the magnitude of two vectors A⃗ and B⃗ and θ is the angle between them.
⇒cosθ=0
⇒θ=cos−1(0)
⇒θ=900
vector product is given by
A×B=∣∣∣∣∣i^−23j^32k^00∣∣∣∣∣
A×B=i^(0–0)−j^(0–0)+k^(−4–9)
A×B=−13k^
or you can do using the properties
i^×j^=j^×j^=k^×k^=0
i^×j^=k^,j^×k^=i^andk^×i^=j^(clockwise)
i^×k^=−j^,j^×i^=−k^andk^×j^=−i^(anticlockwise)
A⃗ ×B⃗ =(−2i^+3j^)×(3i^+2j^)
⇒A×B=−4(i^×j^)+9(j^×i^)
∴A×B=−4k^+9(−k^)=−13k^
Answer:
Hope you like the answer
Explanation:
[A=2i+21+3k] ▪︎[B=3i-21-4k]
= ( 2×2)-(21×12)-(4×3) [ identity, i.i = j.j = k.k = 1 ]
= 4-144-12
= -140-12
= -152