PQRS is a rhombus. Given that $overrightarrow{PQ}=a$ and $ overrightarrow{QR}=b$.
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PQRS is a rhombus. Given that $overrightarrow{PQ}=a$ and $
overrightarrow{QR}=b$.
(a) express the vectors $overrightarrow{PR}$ and $
overrightarrow{QS}$ in terms of $a$ and $b$
(b) hence show that the diagonals in a rhombus intersect at right
angles.
My Attempt:
(a) $overrightarrow{PR}=a+b$ and $overrightarrow{QS}=b-a
$
I have no idea how to work out part (b). Any help is appreciated.
vectors
asked Nov 14 at 3:50
Meghan C
19927
add a comment |
up vote
0
down vote
favorite
PQRS is a rhombus. Given that $overrightarrow{PQ}=a$ and $
overrightarrow{QR}=b$.
(a) express the vectors $overrightarrow{PR}$ and $
overrightarrow{QS}$ in terms of $a$ and $b$
(b) hence show that the diagonals in a rhombus intersect at right
angles.
My Attempt:
(a) $overrightarrow{PR}=a+b$ and $overrightarrow{QS}=b-a
$
I have no idea how to work out part (b). Any help is appreciated.
vectors
asked Nov 14 at 3:50
Meghan C
19927
now show that $overrightarrow{PR} cdot overrightarrow{QS}=0$
– Vasya
Nov 14 at 4:00
@Vasya, Please check my comment below.
– Meghan C
Nov 14 at 4:22
you'll get $-overrightarrow{a}overrightarrow{a}+overrightarrow{b}overrightarrow{b}=0$ because both vectors have the same magnitude (this is rhombus).
– Vasya
Nov 14 at 4:32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
PQRS is a rhombus. Given that $overrightarrow{PQ}=a$ and $
overrightarrow{QR}=b$.
(a) express the vectors $overrightarrow{PR}$ and $
overrightarrow{QS}$ in terms of $a$ and $b$
(b) hence show that the diagonals in a rhombus intersect at right
angles.
My Attempt:
(a) $overrightarrow{PR}=a+b$ and $overrightarrow{QS}=b-a
$
I have no idea how to work out part (b). Any help is appreciated.
vectors
asked Nov 14 at 3:50
Meghan C
19927
PQRS is a rhombus. Given that $overrightarrow{PQ}=a$ and $
overrightarrow{QR}=b$.
(a) express the vectors $overrightarrow{PR}$ and $
overrightarrow{QS}$ in terms of $a$ and $b$
(b) hence show that the diagonals in a rhombus intersect at right
angles.
My Attempt:
(a) $overrightarrow{PR}=a+b$ and $overrightarrow{QS}=b-a
$
I have no idea how to work out part (b). Any help is appreciated.
vectors
vectors
asked Nov 14 at 3:50
Meghan C
19927
asked Nov 14 at 3:50
Meghan C
19927
asked Nov 14 at 3:50
Meghan C
19927
asked Nov 14 at 3:50
Meghan C
19927
asked Nov 14 at 3:50
Meghan C
19927
19927
now show that $overrightarrow{PR} cdot overrightarrow{QS}=0$
– Vasya
Nov 14 at 4:00
@Vasya, Please check my comment below.
– Meghan C
Nov 14 at 4:22
you'll get $-overrightarrow{a}overrightarrow{a}+overrightarrow{b}overrightarrow{b}=0$ because both vectors have the same magnitude (this is rhombus).
– Vasya
Nov 14 at 4:32
add a comment |
now show that $overrightarrow{PR} cdot overrightarrow{QS}=0$
– Vasya
Nov 14 at 4:00
@Vasya, Please check my comment below.
– Meghan C
Nov 14 at 4:22
you'll get $-overrightarrow{a}overrightarrow{a}+overrightarrow{b}overrightarrow{b}=0$ because both vectors have the same magnitude (this is rhombus).
– Vasya
Nov 14 at 4:32
now show that $overrightarrow{PR} cdot overrightarrow{QS}=0$
– Vasya
Nov 14 at 4:00
now show that $overrightarrow{PR} cdot overrightarrow{QS}=0$
– Vasya
Nov 14 at 4:00
@Vasya, Please check my comment below.
– Meghan C
Nov 14 at 4:22
@Vasya, Please check my comment below.
– Meghan C
Nov 14 at 4:22
you'll get $-overrightarrow{a}overrightarrow{a}+overrightarrow{b}overrightarrow{b}=0$ because both vectors have the same magnitude (this is rhombus).
– Vasya
Nov 14 at 4:32
you'll get $-overrightarrow{a}overrightarrow{a}+overrightarrow{b}overrightarrow{b}=0$ because both vectors have the same magnitude (this is rhombus).
– Vasya
Nov 14 at 4:32
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Just take the dot product of diagonals, that is 0( because b=a), hence proved.
Hope it helps.
answered Nov 14 at 4:00
Crazy for maths
4948
I am stuck where (a+b).(b-a)=0, how do you proceed from there? Please provide a detailed explanation, because I do not understand how b = a will help here.
– Meghan C
Nov 14 at 4:21
Multiplication of vectors is like numbers and all algebraic identities hold because of commutativity of dot product
– Crazy for maths
Nov 14 at 4:29
$(vec a+vec b)(vec b-vec a)=vec avec b-vec avec a+vec bvec b-vec bvec a$. For dot product $vec avec b=vec bvec a$ and $vec a vec a=a^2$
– Andrei
Nov 14 at 4:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Just take the dot product of diagonals, that is 0( because b=a), hence proved.
Hope it helps.
answered Nov 14 at 4:00
Crazy for maths
4948
I am stuck where (a+b).(b-a)=0, how do you proceed from there? Please provide a detailed explanation, because I do not understand how b = a will help here.
– Meghan C
Nov 14 at 4:21
Multiplication of vectors is like numbers and all algebraic identities hold because of commutativity of dot product
– Crazy for maths
Nov 14 at 4:29
$(vec a+vec b)(vec b-vec a)=vec avec b-vec avec a+vec bvec b-vec bvec a$. For dot product $vec avec b=vec bvec a$ and $vec a vec a=a^2$
– Andrei
Nov 14 at 4:29
add a comment |
up vote
0
down vote
Just take the dot product of diagonals, that is 0( because b=a), hence proved.
Hope it helps.
answered Nov 14 at 4:00
Crazy for maths
4948
I am stuck where (a+b).(b-a)=0, how do you proceed from there? Please provide a detailed explanation, because I do not understand how b = a will help here.
– Meghan C
Nov 14 at 4:21
Multiplication of vectors is like numbers and all algebraic identities hold because of commutativity of dot product
– Crazy for maths
Nov 14 at 4:29
$(vec a+vec b)(vec b-vec a)=vec avec b-vec avec a+vec bvec b-vec bvec a$. For dot product $vec avec b=vec bvec a$ and $vec a vec a=a^2$
– Andrei
Nov 14 at 4:29
add a comment |
up vote
0
down vote
up vote
0
down vote
Just take the dot product of diagonals, that is 0( because b=a), hence proved.
Hope it helps.
answered Nov 14 at 4:00
Crazy for maths
4948
Just take the dot product of diagonals, that is 0( because b=a), hence proved.
Hope it helps.
answered Nov 14 at 4:00
Crazy for maths
4948
answered Nov 14 at 4:00
Crazy for maths
4948
answered Nov 14 at 4:00
Crazy for maths
4948
answered Nov 14 at 4:00
Crazy for maths
4948
4948
I am stuck where (a+b).(b-a)=0, how do you proceed from there? Please provide a detailed explanation, because I do not understand how b = a will help here.
– Meghan C
Nov 14 at 4:21
Multiplication of vectors is like numbers and all algebraic identities hold because of commutativity of dot product
– Crazy for maths
Nov 14 at 4:29
$(vec a+vec b)(vec b-vec a)=vec avec b-vec avec a+vec bvec b-vec bvec a$. For dot product $vec avec b=vec bvec a$ and $vec a vec a=a^2$
– Andrei
Nov 14 at 4:29
add a comment |
I am stuck where (a+b).(b-a)=0, how do you proceed from there? Please provide a detailed explanation, because I do not understand how b = a will help here.
– Meghan C
Nov 14 at 4:21
Multiplication of vectors is like numbers and all algebraic identities hold because of commutativity of dot product
– Crazy for maths
Nov 14 at 4:29
$(vec a+vec b)(vec b-vec a)=vec avec b-vec avec a+vec bvec b-vec bvec a$. For dot product $vec avec b=vec bvec a$ and $vec a vec a=a^2$
– Andrei
Nov 14 at 4:29
I am stuck where (a+b).(b-a)=0, how do you proceed from there? Please provide a detailed explanation, because I do not understand how b = a will help here.
– Meghan C
Nov 14 at 4:21
I am stuck where (a+b).(b-a)=0, how do you proceed from there? Please provide a detailed explanation, because I do not understand how b = a will help here.
– Meghan C
Nov 14 at 4:21
Multiplication of vectors is like numbers and all algebraic identities hold because of commutativity of dot product
– Crazy for maths
Nov 14 at 4:29
Multiplication of vectors is like numbers and all algebraic identities hold because of commutativity of dot product
– Crazy for maths
Nov 14 at 4:29
$(vec a+vec b)(vec b-vec a)=vec avec b-vec avec a+vec bvec b-vec bvec a$. For dot product $vec avec b=vec bvec a$ and $vec a vec a=a^2$
– Andrei
Nov 14 at 4:29
$(vec a+vec b)(vec b-vec a)=vec avec b-vec avec a+vec bvec b-vec bvec a$. For dot product $vec avec b=vec bvec a$ and $vec a vec a=a^2$
– Andrei
Nov 14 at 4:29
add a comment |
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now show that $overrightarrow{PR} cdot overrightarrow{QS}=0$
– Vasya
Nov 14 at 4:00
@Vasya, Please check my comment below.
– Meghan C
Nov 14 at 4:22
you'll get $-overrightarrow{a}overrightarrow{a}+overrightarrow{b}overrightarrow{b}=0$ because both vectors have the same magnitude (this is rhombus).
– Vasya
Nov 14 at 4:32