How to find out the modulus of vectors in this question (as seen in the image):How to find the “average” direction of a set of vectors?Flux intergrals/Finding normals and their dot productsFind the equation of a plane when it passes through two points and parallel to two vectorsFind the direction in which the function $f(x, y)$ = $x^3y − x^2y^4$ decreases fastest at the point (5, −4).How to find the vector formula for the bisector of given two vectors?How to find the dot product inside of a squareFind the norm/magnitude of 2 cross products of vectorsHow to find the plane with given 2 vectors?How do I work with $2$D vectors where the origin is not at $(0, 0)$?For the question attached, how would you find the values of the vectors BA and BC?
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How to find out the modulus of vectors in this question (as seen in the image):
How to find the “average” direction of a set of vectors?Flux intergrals/Finding normals and their dot productsFind the equation of a plane when it passes through two points and parallel to two vectorsFind the direction in which the function $f(x, y)$ = $x^3y − x^2y^4$ decreases fastest at the point (5, −4).How to find the vector formula for the bisector of given two vectors?How to find the dot product inside of a squareFind the norm/magnitude of 2 cross products of vectorsHow to find the plane with given 2 vectors?How do I work with $2$D vectors where the origin is not at $(0, 0)$?For the question attached, how would you find the values of the vectors BA and BC?
$begingroup$
A problem I'm facing involves finding out the modulus of vectors. I understand that the modulus of a vector is the square of both the numbers and then the root of it to find the length. I was thinking for part (a) the answer would be 3 + (2*3) which is 9, and then the square root of this which will be 3. I used the same method to do part (b). However, I'm not sure whether what I have done is right, and I am stuck on part (c).
vectors
asked 2 days ago
V11V11
196
$endgroup$
add a comment |
$begingroup$
A problem I'm facing involves finding out the modulus of vectors. I understand that the modulus of a vector is the square of both the numbers and then the root of it to find the length. I was thinking for part (a) the answer would be 3 + (2*3) which is 9, and then the square root of this which will be 3. I used the same method to do part (b). However, I'm not sure whether what I have done is right, and I am stuck on part (c).
vectors
asked 2 days ago
V11V11
196
$endgroup$
add a comment |
$begingroup$
A problem I'm facing involves finding out the modulus of vectors. I understand that the modulus of a vector is the square of both the numbers and then the root of it to find the length. I was thinking for part (a) the answer would be 3 + (2*3) which is 9, and then the square root of this which will be 3. I used the same method to do part (b). However, I'm not sure whether what I have done is right, and I am stuck on part (c).
vectors
asked 2 days ago
V11V11
196
$endgroup$
A problem I'm facing involves finding out the modulus of vectors. I understand that the modulus of a vector is the square of both the numbers and then the root of it to find the length. I was thinking for part (a) the answer would be 3 + (2*3) which is 9, and then the square root of this which will be 3. I used the same method to do part (b). However, I'm not sure whether what I have done is right, and I am stuck on part (c).
vectors
vectors
asked 2 days ago
V11V11
196
asked 2 days ago
V11V11
196
asked 2 days ago
V11V11
196
asked 2 days ago
V11V11
196
asked 2 days ago
V11V11
196
196
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $$b=2a$$ then we get $$|a+b|=|a+2a|=|3a|=3|a|$$
If $$b=-2a$$ then we get $$3|-a|=3|a|$$
If $a$ is perpendicular to $b$ then $$|a+b|^2=|a|^2+|b|^2=9+16=25$$ and $$|a+b|$$ gives $5$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
$begingroup$
Oh, thank you for your answer. However, just to make sure I understood the concept, I had a doubt. If b = -3a, wouldn't it become |a-3a| = |-2a|, which would then become -2|a|, resulting in -6 as the answer. The answer says 6 (I am not so sure of the rules of vectors like what would happen if the number is negative). Also, is there any rule that allows the two variables inside the modulus to be separated as done for part (c)? Thanks once again.
$endgroup$
– V11
2 days ago
$begingroup$
Note that $$|a-3a|=|-2a|=|-2||a|=2|a|$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
note that $$(a+b)cdot (a+b)=(a+b)^2=|a+b|^2$$ if $$cdot$$ means dot product.
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Oh yes I get the second part. However, I didn't exactly understand how |−2||𝑎| would become 2|𝑎|.
$endgroup$
– V11
2 days ago
$begingroup$
We have this law also with real numbers and $$|-2|=2$$, by definition is it $$-(-2)=2$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
|
show 3 more comments
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $$b=2a$$ then we get $$|a+b|=|a+2a|=|3a|=3|a|$$
If $$b=-2a$$ then we get $$3|-a|=3|a|$$
If $a$ is perpendicular to $b$ then $$|a+b|^2=|a|^2+|b|^2=9+16=25$$ and $$|a+b|$$ gives $5$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
$begingroup$
Oh, thank you for your answer. However, just to make sure I understood the concept, I had a doubt. If b = -3a, wouldn't it become |a-3a| = |-2a|, which would then become -2|a|, resulting in -6 as the answer. The answer says 6 (I am not so sure of the rules of vectors like what would happen if the number is negative). Also, is there any rule that allows the two variables inside the modulus to be separated as done for part (c)? Thanks once again.
$endgroup$
– V11
2 days ago
$begingroup$
Note that $$|a-3a|=|-2a|=|-2||a|=2|a|$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
note that $$(a+b)cdot (a+b)=(a+b)^2=|a+b|^2$$ if $$cdot$$ means dot product.
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Oh yes I get the second part. However, I didn't exactly understand how |−2||𝑎| would become 2|𝑎|.
$endgroup$
– V11
2 days ago
$begingroup$
We have this law also with real numbers and $$|-2|=2$$, by definition is it $$-(-2)=2$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
|
show 3 more comments
$begingroup$
If $$b=2a$$ then we get $$|a+b|=|a+2a|=|3a|=3|a|$$
If $$b=-2a$$ then we get $$3|-a|=3|a|$$
If $a$ is perpendicular to $b$ then $$|a+b|^2=|a|^2+|b|^2=9+16=25$$ and $$|a+b|$$ gives $5$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
$begingroup$
Oh, thank you for your answer. However, just to make sure I understood the concept, I had a doubt. If b = -3a, wouldn't it become |a-3a| = |-2a|, which would then become -2|a|, resulting in -6 as the answer. The answer says 6 (I am not so sure of the rules of vectors like what would happen if the number is negative). Also, is there any rule that allows the two variables inside the modulus to be separated as done for part (c)? Thanks once again.
$endgroup$
– V11
2 days ago
$begingroup$
Note that $$|a-3a|=|-2a|=|-2||a|=2|a|$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
note that $$(a+b)cdot (a+b)=(a+b)^2=|a+b|^2$$ if $$cdot$$ means dot product.
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Oh yes I get the second part. However, I didn't exactly understand how |−2||𝑎| would become 2|𝑎|.
$endgroup$
– V11
2 days ago
$begingroup$
We have this law also with real numbers and $$|-2|=2$$, by definition is it $$-(-2)=2$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
|
show 3 more comments
$begingroup$
If $$b=2a$$ then we get $$|a+b|=|a+2a|=|3a|=3|a|$$
If $$b=-2a$$ then we get $$3|-a|=3|a|$$
If $a$ is perpendicular to $b$ then $$|a+b|^2=|a|^2+|b|^2=9+16=25$$ and $$|a+b|$$ gives $5$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
If $$b=2a$$ then we get $$|a+b|=|a+2a|=|3a|=3|a|$$
If $$b=-2a$$ then we get $$3|-a|=3|a|$$
If $a$ is perpendicular to $b$ then $$|a+b|^2=|a|^2+|b|^2=9+16=25$$ and $$|a+b|$$ gives $5$
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
77.7k42866
$begingroup$
Oh, thank you for your answer. However, just to make sure I understood the concept, I had a doubt. If b = -3a, wouldn't it become |a-3a| = |-2a|, which would then become -2|a|, resulting in -6 as the answer. The answer says 6 (I am not so sure of the rules of vectors like what would happen if the number is negative). Also, is there any rule that allows the two variables inside the modulus to be separated as done for part (c)? Thanks once again.
$endgroup$
– V11
2 days ago
$begingroup$
Note that $$|a-3a|=|-2a|=|-2||a|=2|a|$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
note that $$(a+b)cdot (a+b)=(a+b)^2=|a+b|^2$$ if $$cdot$$ means dot product.
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Oh yes I get the second part. However, I didn't exactly understand how |−2||𝑎| would become 2|𝑎|.
$endgroup$
– V11
2 days ago
$begingroup$
We have this law also with real numbers and $$|-2|=2$$, by definition is it $$-(-2)=2$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
|
show 3 more comments
$begingroup$
Oh, thank you for your answer. However, just to make sure I understood the concept, I had a doubt. If b = -3a, wouldn't it become |a-3a| = |-2a|, which would then become -2|a|, resulting in -6 as the answer. The answer says 6 (I am not so sure of the rules of vectors like what would happen if the number is negative). Also, is there any rule that allows the two variables inside the modulus to be separated as done for part (c)? Thanks once again.
$endgroup$
– V11
2 days ago
$begingroup$
Note that $$|a-3a|=|-2a|=|-2||a|=2|a|$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
note that $$(a+b)cdot (a+b)=(a+b)^2=|a+b|^2$$ if $$cdot$$ means dot product.
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Oh yes I get the second part. However, I didn't exactly understand how |−2||𝑎| would become 2|𝑎|.
$endgroup$
– V11
2 days ago
$begingroup$
We have this law also with real numbers and $$|-2|=2$$, by definition is it $$-(-2)=2$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Oh, thank you for your answer. However, just to make sure I understood the concept, I had a doubt. If b = -3a, wouldn't it become |a-3a| = |-2a|, which would then become -2|a|, resulting in -6 as the answer. The answer says 6 (I am not so sure of the rules of vectors like what would happen if the number is negative). Also, is there any rule that allows the two variables inside the modulus to be separated as done for part (c)? Thanks once again.
$endgroup$
– V11
2 days ago
$begingroup$
Oh, thank you for your answer. However, just to make sure I understood the concept, I had a doubt. If b = -3a, wouldn't it become |a-3a| = |-2a|, which would then become -2|a|, resulting in -6 as the answer. The answer says 6 (I am not so sure of the rules of vectors like what would happen if the number is negative). Also, is there any rule that allows the two variables inside the modulus to be separated as done for part (c)? Thanks once again.
$endgroup$
– V11
2 days ago
$begingroup$
Note that $$|a-3a|=|-2a|=|-2||a|=2|a|$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Note that $$|a-3a|=|-2a|=|-2||a|=2|a|$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
note that $$(a+b)cdot (a+b)=(a+b)^2=|a+b|^2$$ if $$cdot$$ means dot product.
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
note that $$(a+b)cdot (a+b)=(a+b)^2=|a+b|^2$$ if $$cdot$$ means dot product.
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
Oh yes I get the second part. However, I didn't exactly understand how |−2||𝑎| would become 2|𝑎|.
$endgroup$
– V11
2 days ago
$begingroup$
Oh yes I get the second part. However, I didn't exactly understand how |−2||𝑎| would become 2|𝑎|.
$endgroup$
– V11
2 days ago
$begingroup$
We have this law also with real numbers and $$|-2|=2$$, by definition is it $$-(-2)=2$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
$begingroup$
We have this law also with real numbers and $$|-2|=2$$, by definition is it $$-(-2)=2$$
$endgroup$
– Dr. Sonnhard Graubner
2 days ago
|
show 3 more comments
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