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Vectors , finding coefficient
Find coordinates of point that satisfy given conditionsFinding an angle between two vectors without a calculatorUse of determinants for vectors - what is the intuition behind it?Coordinate Systems Transformation(Rectangular to Cylindrical)Finding three new vectors pointing towards the vertices of a regular tetrahedron, with one vector given.Area of paralellogramHow to solve a matrix multiplication that results in zero? F*x = 0Check if three row vectors are linearly dependent or independentMinimization of the distance between 2 vectorsFinding Matrix given an Equation
$begingroup$
if (x,y,z) unequal to (0,0,0) and (i + j + 3k)x + ( 3i - 3j +k)y + (-4i + 5j)z = a(xi + yj + zk), then the value of a is
Attempt:
I wrote
$$(x+3y-4z) i + (x-3y+5z) j + (3x+y) k = a (x i + y j + z k)$$
Which gave me
$$
x+3y-4z=ax\
x-3y+5z=ay\
3x+y=az
$$
Now these are 4 unknown and only 3 equations.
Also, i dont know determinant , so please suggest me another way to solve it without using determinants.
vectors vector-analysis
edited Mar 13 at 8:28
Arthur
118k7118201
asked Mar 13 at 8:26
BadguyBadguy
113
$endgroup$
add a comment |
$begingroup$
if (x,y,z) unequal to (0,0,0) and (i + j + 3k)x + ( 3i - 3j +k)y + (-4i + 5j)z = a(xi + yj + zk), then the value of a is
Attempt:
I wrote
$$(x+3y-4z) i + (x-3y+5z) j + (3x+y) k = a (x i + y j + z k)$$
Which gave me
$$
x+3y-4z=ax\
x-3y+5z=ay\
3x+y=az
$$
Now these are 4 unknown and only 3 equations.
Also, i dont know determinant , so please suggest me another way to solve it without using determinants.
vectors vector-analysis
edited Mar 13 at 8:28
Arthur
118k7118201
asked Mar 13 at 8:26
BadguyBadguy
113
$endgroup$
$begingroup$
Well, my first thought is that you should use the determinant :)
$endgroup$
– Matti P.
Mar 13 at 8:31
1
$begingroup$
You have 4 unknowns and 3 equations, so it's clearly underdetermined. But that's a feature of the original problem as well: if you have one solution, then multiplying each of $x, y, z$ with any non-zero number gives you another solution. So I would just pretend that one of the unknowns (say, $x$) is actually known and solve for the rest.
$endgroup$
– Arthur
Mar 13 at 8:31
$begingroup$
Actually, I would write the equations as $$ left{ beginarrayrrrc (1-a)x&+3y&-4z&=0\ x&-(3+a)y&+5z&=0\ 3x&+y&-az& = 0 endarray right. $$ or in the matrix form $$ left[ beginarrayrrrc (1-a)&3&-4\ 1&-(3+a)&5\ 3&1&-a endarray right] left[ beginarrayc x \ y \ z endarray right] = left[ beginarrayc 0 \ 0 \ 0 endarray right] $$ Now the question becomes: Is there a non-trivial solution to this equation?
$endgroup$
– Matti P.
Mar 13 at 8:37
add a comment |
$begingroup$
if (x,y,z) unequal to (0,0,0) and (i + j + 3k)x + ( 3i - 3j +k)y + (-4i + 5j)z = a(xi + yj + zk), then the value of a is
Attempt:
I wrote
$$(x+3y-4z) i + (x-3y+5z) j + (3x+y) k = a (x i + y j + z k)$$
Which gave me
$$
x+3y-4z=ax\
x-3y+5z=ay\
3x+y=az
$$
Now these are 4 unknown and only 3 equations.
Also, i dont know determinant , so please suggest me another way to solve it without using determinants.
vectors vector-analysis
edited Mar 13 at 8:28
Arthur
118k7118201
asked Mar 13 at 8:26
BadguyBadguy
113
$endgroup$
if (x,y,z) unequal to (0,0,0) and (i + j + 3k)x + ( 3i - 3j +k)y + (-4i + 5j)z = a(xi + yj + zk), then the value of a is
Attempt:
I wrote
$$(x+3y-4z) i + (x-3y+5z) j + (3x+y) k = a (x i + y j + z k)$$
Which gave me
$$
x+3y-4z=ax\
x-3y+5z=ay\
3x+y=az
$$
Now these are 4 unknown and only 3 equations.
Also, i dont know determinant , so please suggest me another way to solve it without using determinants.
vectors vector-analysis
vectors vector-analysis
edited Mar 13 at 8:28
Arthur
118k7118201
asked Mar 13 at 8:26
BadguyBadguy
113
edited Mar 13 at 8:28
Arthur
118k7118201
asked Mar 13 at 8:26
BadguyBadguy
113
edited Mar 13 at 8:28
Arthur
118k7118201
edited Mar 13 at 8:28
Arthur
118k7118201
edited Mar 13 at 8:28
Arthur
118k7118201
118k7118201
asked Mar 13 at 8:26
BadguyBadguy
113
asked Mar 13 at 8:26
BadguyBadguy
113
asked Mar 13 at 8:26
BadguyBadguy
113
113
$begingroup$
Well, my first thought is that you should use the determinant :)
$endgroup$
– Matti P.
Mar 13 at 8:31
1
$begingroup$
You have 4 unknowns and 3 equations, so it's clearly underdetermined. But that's a feature of the original problem as well: if you have one solution, then multiplying each of $x, y, z$ with any non-zero number gives you another solution. So I would just pretend that one of the unknowns (say, $x$) is actually known and solve for the rest.
$endgroup$
– Arthur
Mar 13 at 8:31
$begingroup$
Actually, I would write the equations as $$ left{ beginarrayrrrc (1-a)x&+3y&-4z&=0\ x&-(3+a)y&+5z&=0\ 3x&+y&-az& = 0 endarray right. $$ or in the matrix form $$ left[ beginarrayrrrc (1-a)&3&-4\ 1&-(3+a)&5\ 3&1&-a endarray right] left[ beginarrayc x \ y \ z endarray right] = left[ beginarrayc 0 \ 0 \ 0 endarray right] $$ Now the question becomes: Is there a non-trivial solution to this equation?
$endgroup$
– Matti P.
Mar 13 at 8:37
add a comment |
$begingroup$
Well, my first thought is that you should use the determinant :)
$endgroup$
– Matti P.
Mar 13 at 8:31
1
$begingroup$
You have 4 unknowns and 3 equations, so it's clearly underdetermined. But that's a feature of the original problem as well: if you have one solution, then multiplying each of $x, y, z$ with any non-zero number gives you another solution. So I would just pretend that one of the unknowns (say, $x$) is actually known and solve for the rest.
$endgroup$
– Arthur
Mar 13 at 8:31
$begingroup$
Actually, I would write the equations as $$ left{ beginarrayrrrc (1-a)x&+3y&-4z&=0\ x&-(3+a)y&+5z&=0\ 3x&+y&-az& = 0 endarray right. $$ or in the matrix form $$ left[ beginarrayrrrc (1-a)&3&-4\ 1&-(3+a)&5\ 3&1&-a endarray right] left[ beginarrayc x \ y \ z endarray right] = left[ beginarrayc 0 \ 0 \ 0 endarray right] $$ Now the question becomes: Is there a non-trivial solution to this equation?
$endgroup$
– Matti P.
Mar 13 at 8:37
$begingroup$
Well, my first thought is that you should use the determinant :)
$endgroup$
– Matti P.
Mar 13 at 8:31
$begingroup$
Well, my first thought is that you should use the determinant :)
$endgroup$
– Matti P.
Mar 13 at 8:31
1
1
$begingroup$
You have 4 unknowns and 3 equations, so it's clearly underdetermined. But that's a feature of the original problem as well: if you have one solution, then multiplying each of $x, y, z$ with any non-zero number gives you another solution. So I would just pretend that one of the unknowns (say, $x$) is actually known and solve for the rest.
$endgroup$
– Arthur
Mar 13 at 8:31
$begingroup$
You have 4 unknowns and 3 equations, so it's clearly underdetermined. But that's a feature of the original problem as well: if you have one solution, then multiplying each of $x, y, z$ with any non-zero number gives you another solution. So I would just pretend that one of the unknowns (say, $x$) is actually known and solve for the rest.
$endgroup$
– Arthur
Mar 13 at 8:31
$begingroup$
Actually, I would write the equations as $$ left{ beginarrayrrrc (1-a)x&+3y&-4z&=0\ x&-(3+a)y&+5z&=0\ 3x&+y&-az& = 0 endarray right. $$ or in the matrix form $$ left[ beginarrayrrrc (1-a)&3&-4\ 1&-(3+a)&5\ 3&1&-a endarray right] left[ beginarrayc x \ y \ z endarray right] = left[ beginarrayc 0 \ 0 \ 0 endarray right] $$ Now the question becomes: Is there a non-trivial solution to this equation?
$endgroup$
– Matti P.
Mar 13 at 8:37
$begingroup$
Actually, I would write the equations as $$ left{ beginarrayrrrc (1-a)x&+3y&-4z&=0\ x&-(3+a)y&+5z&=0\ 3x&+y&-az& = 0 endarray right. $$ or in the matrix form $$ left[ beginarrayrrrc (1-a)&3&-4\ 1&-(3+a)&5\ 3&1&-a endarray right] left[ beginarrayc x \ y \ z endarray right] = left[ beginarrayc 0 \ 0 \ 0 endarray right] $$ Now the question becomes: Is there a non-trivial solution to this equation?
$endgroup$
– Matti P.
Mar 13 at 8:37
add a comment |
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$begingroup$
Well, my first thought is that you should use the determinant :)
$endgroup$
– Matti P.
Mar 13 at 8:31
1
$begingroup$
You have 4 unknowns and 3 equations, so it's clearly underdetermined. But that's a feature of the original problem as well: if you have one solution, then multiplying each of $x, y, z$ with any non-zero number gives you another solution. So I would just pretend that one of the unknowns (say, $x$) is actually known and solve for the rest.
$endgroup$
– Arthur
Mar 13 at 8:31
$begingroup$
Actually, I would write the equations as $$ left{ beginarrayrrrc (1-a)x&+3y&-4z&=0\ x&-(3+a)y&+5z&=0\ 3x&+y&-az& = 0 endarray right. $$ or in the matrix form $$ left[ beginarrayrrrc (1-a)&3&-4\ 1&-(3+a)&5\ 3&1&-a endarray right] left[ beginarrayc x \ y \ z endarray right] = left[ beginarrayc 0 \ 0 \ 0 endarray right] $$ Now the question becomes: Is there a non-trivial solution to this equation?
$endgroup$
– Matti P.
Mar 13 at 8:37