How to show these vectors are perpendicular

[Guest]

Show that these vectors are perpendicular to each other.
A = i + j + k
B = 2i -3j + k
C = 4i + j -5k
Thanks

[Math Tutor]

Prove that the scalar(dot) product of the vectors is 0.
The scalar product of A.B = 1 * 2 + 1 * (-3) + 1 * 1 = 0.

[Math Tutor]

How will you show that B and C are perpendicular.

[HallsofIvy]

I won't! But you can by doing exactly what Math Tutor said. Do you know what the "scalar product" or "dot product" of two vectors is?

[Guest]

[tex]\vec{A}\cdot\vec{B}=
(1\vec{i}+1\vec{j}+1\vec{k})\cdot(2\vec{i}+(-3)\vec{j}+1\vec{k})=
1\vec{i}\cdot2\vec{i}+1\vec{j}\cdot2\vec{i}+1\vec{k}\cdot2\vec{i}+1\vec{i}\cdot(-3)\vec{j}+1\vec{j}\cdot(-3)\vec{j}+1\vec{k}\cdot(-3)\vec{j}+1\vec{i}\cdot1\vec{k}+1\vec{j}\cdot1\vec{k}+1\vec{k}\cdot1\vec{k}=
(1\cdot2)( \vec{i} \cdot \vec{i} )+(1\cdot2)(\vec{j}\cdot\vec{i})+(1\cdot2)(\vec{k}\cdot\vec{i})+(1\cdot(-3))(\vec{i}\cdot\vec{j})+(1\cdot(-3))(\vec{j}\cdot\vec{j})+(1\cdot(-3))(\vec{k}\cdot\vec{j})+(1\cdot1)(\vec{i}\cdot\vec{k})+(1\cdot1)(\vec{j}\cdot\vec{k})+(1\cdot1)(\vec{k}\cdot\vec{k})=
2\cdot1+2\cdot0+2\cdot0+(-3)\cdot0+(-3)\cdot1+(-3)\cdot0+1\cdot0+1\cdot0+1\cdot1=
2+0+0+0+(-3)+0+0+0+1=
0[/tex]
[tex]\vec{B}\cdot\vec{C}=
(2\vec{i}+(-3)\vec{j}+1\vec{k})\cdot(4\vec{i}+1\vec{j}+(-5)\vec{k})=
2\vec{i}\cdot4\vec{i}+(-3)\vec{j}\cdot4\vec{i}+1\vec{k}\cdot4\vec{i}+2\vec{i}\cdot1\vec{j}+(-3)\vec{j}\cdot1\vec{j}+1\vec{k}\cdot1\vec{j}+2\vec{i}\cdot(-5)\vec{k}+(-3)\vec{j}\cdot(-5)\vec(k)+1\vec{k}\cdot(-5)\vec{k}=
(2\cdot4)(\vec{i}\cdot\vec{i})+((-3)\cdot4)(\vec{j}\cdot\vec{i})+(1\cdot4)(\vec{k}\cdot\vec{i})+(2\cdot1)(\vec{i}\cdot\vec{j})+((-3)\cdot1)(\vec{j}\cdot\vec{j})+(1\cdot1)(\vec{k}\cdot\vec{j})+(2\cdot(-5))(\vec{i}\cdot\vec{k})+((-3)\cdot(-5))(\vec{j}\cdot\vec{k})+(1\cdot(-5))(\vec{k}\cdot\vec{k})=
8\cdot1+(-12)\cdot0+4\cdot0+2\cdot0+(-3)\cdot1+1\cdot0+(-10)\cdot0+15\cdot0+(-5)\cdot1=
8+0+0+0+(-3)+0+0+0+(-5)=
0[/tex]
[tex]\vec{C}\cdot\vec{A}=
(4\vec{i}+1\vec{j}+(-5)\vec{k})\cdot(1\vec{i}+1\vec{j}+1\vec{k})=
4\vec{i}\cdot1\vec{i}+1\vec{j}\cdot1\vec{i}+(-5)\vec{k}\cdot1\vec{i}+4\vec{i}\cdot1\vec{j}+1\vec{j}\cdot1\vec{j}+(-5)\vec{k}\cdot1\vec{j}+4\vec{i}\cdot1\vec{k}+1\vec{j}\cdot1\vec(k)+(-5)\vec{k}\cdot1\vec{k}=
(4\cdot1)(\vec{i}\cdot\vec{i})+(1\cdot1)(\vec{j}\cdot\vec{i})+((-5)\cdot1)(\vec{k}\cdot\vec{i})+(4\cdot1)(\vec{i}\cdot\vec{j})+(1\cdot1)(\vec{j}\cdot\vec{j})+((-5)\cdot1)(\vec{k}\cdot\vec{j})+(4\cdot1)(\vec{i}\cdot\vec{k})+(1\cdot1)(\vec{j}\cdot\vec{k})+((-5)\cdot1)(\vec{k}\cdot\vec{k})=
4\cdot1+1\cdot0+(-5)\cdot0+4\cdot0+1\cdot1+(-5)\cdot0+4\cdot0+1\cdot0+(-5)\cdot1=
4+0+0+0+1+0+0+0+(-5)=
0[/tex]

[Guest]

To show that the vectors A, B, and C are perpendicular to each other, we need to verify that their dot products are zero.
Given:
A=i+j+k
B=2i−3j+k
C=4i+j−5k
Calculate A⋅B:
A⋅B=(1)(2)+(1)(−3)+(1)(1)=2−3+1=0
Calculate A⋅C:
A⋅C=(1)(4)+(1)(1)+(1)(−5)=4+1−5=0
Calculate B⋅C:
B⋅C=(2)(4)+(−3)(1)+(1)(−5)=8−3−5=0
Since A⋅B=0, A⋅C=0 and B⋅C=0, the vectors A, B, and C are all perpendicular to each other.

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