Question

Find a unit vector perpendicular to each of the vectors = 5i +6j-2k and b = 7i +6j+2k​

Answer 1

Answer:  To find a unit vector that is perpendicular to the vectors a = 5i + 6j - 2k and b = 7i + 6j + 2k, we can use two methods:

Cross product method:

(a x b) = (5i + 6j - 2k) x (7i + 6j + 2k) = (6k - 12i)i + (-30j)j + (12k + 10i)k

Normalize the above cross product by dividing it by its magnitude.

Dot product method:

Take any linear combination of the two vectors c = xa + yb, such that c . a = 0 and c . b = 0, this will give you the coefficients x and y, and you can find the vector c by plugging these values into the linear combination. Both the cross-product and the dot-product method can be used to find a vector perpendicular to two given vectors. The cross-product method is useful when the vectors are in 3D space and the dot-product method is useful when the vectors are in 2D space.

Explanation: A unit vector that is perpendicular to the vector a = 5i + 6j - 2k and b = 7i + 6j + 2k can be found by taking the cross product of the two vectors:

(a x b) = (5i + 6j - 2k) x (7i + 6j + 2k) = (6k - 12i)i + (-30j)j + (12k + 10i)k

This vector is not a unit vector, so we need to normalize it to get the unit vector.

To do this, we divide the vector by its magnitude:

||(a x b)|| = sqrt((6k - 12i)^2 + (-30j)^2 + (12k + 10i)^2)

(a x b) / ||(a x b)|| = (6k - 12i) / sqrt(164)i + (-30j) / sqrt(164)j + (12k + 10i) / sqrt(164)k

Therefore, the unit vector that is perpendicular to both a and b is:

(6k - 12i) / sqrt(164)i + (-30j) / sqrt(164)j + (12k + 10i) / sqrt(164)k

Dot product method- If the dot product of two vectors is zero, then the vectors are orthogonal or perpendicular to each other.

So to find a vector that is perpendicular to both a and b, we can take any linear combination of the two vectors and set the dot product of the combination to zero:

c = xa + yb

c . a = x(a . a) + y(b . a) = 0

c . b = x(a . b) + y(b . b) = 0

Solving this system of equations will give us the coefficients x and y, and we can then find the vector c by plugging these values into the linear combination.

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Answer 2

Unit vector perpendicular to the given two vectors is (2/ √97)i + (-5/ √97)j + (2/ √97)k.

To find a unit vector that is perpendicular to two given vectors, we can use the cross product of the two vectors. The cross product of two vectors results in a vector that is perpendicular to both input vectors.

The cross product of the two vectors = 5i +6j-2k and b = 7i +6j+2k is given by:

$(5i +6j-2k) x (7i +6j+2k) = (6k-2j)i + (-5k+7i)j + (6i-5j)k$

Now we will find the magnitude of the cross product

$|(6k-2j)i + (-5k+7i)j + (6i-5j)k|$

=$\sqrt((6k-2j)^2 + (-5k+7i)^2 + (6i-5j)^2)$

= √(36+25+36) = √97

So, the unit vector will be given by:

$((6k-2j)i + (-5k+7i)j + (6i-5j)k)$/ √97

= (2/ √97)i + (-5/ √97)j + (2/ √97)k

So, a unit vector perpendicular to the given two vectors is (2/ √97)i + (-5/ √97)j + (2/ √97)k.

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