Let b=4 i+3 j and c be two vectors perpendicular to each...

Let $b=4 i+3 j$ and $c$ be two vectors perpendicular to each other in $x y$-plane, then the vector in the same plane having projections 1 and 2 along $b$ and $c$ respectively is
(A) $2 i-j$
(B) $-2 i+j$
(C) $2 i+j$
(D) none of these

Key Concepts

Vector Projection

This concept involves determining the component of one vector along the direction of another vector. By using the dot product, which measures the magnitude of one vector in the direction of another, one can compute how much of one vector is 'projected' onto another. This is fundamental in resolving a vector into parts that align with specific directions.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at right angles, meaning their dot product is zero. In many applications, using orthogonal vectors as a basis simplifies computations such as projections and decompositions because the components along each direction do not interfere with one another.

Vector Decomposition

Vector decomposition refers to expressing a vector as a sum of components along chosen directions, often chosen to be mutually perpendicular. This technique is useful in breaking down complex vector relationships into simpler, more manageable parts, particularly when the given directions form an orthogonal basis.

Transcript

00:01 In problem 46 we have the vector b equals 4i plus 3g for example this vector and we have another vector c which is perpendicular to b then for example c would be this direction or in this direction we want to find the vector that is in the same plane having projections in one in the direction of b and the direction of b and the direction of c as 1 and 2 then if we have vector a for example this vector we want to get its projection on b and equated to 1 and its projection on c and equated to 2 let's find this vector a first we get we should get c which is perpendicular to be if we assume that c equal c1 i plus c2 j then the dot product between c and b is 0 then 4 c1 plus 3 c2 equal 0 then we can get that c1 equals minus 3 c2 divided by 4 and we can get a unit vector in the direction of c we can get c head c had equal c divided by the norm of c c equal c1 i plus c2 j divided by the norm of c which is square root of c1 squared plus c2 squared c1 squared plus c2 square let's substitute by c1 equals minus 3 divided by 4 c2 then it equals minus 3 divided by 4 i plus j all multiplied by c2 divided by c2 divided by c2 divided by square root of 3 divided by 4 squared plus 1 multiplied by c2 we can remove c2 with c2 then we have in the direction of c we have ionid vector equals 4 divided by 5 multiplied by minus 3 divided by 4 i plus j it equals 1 divided by 5 multiplied by minus 3 i plus 4g this is ioned vector in the direction of c hat to find a we get 2 multiplied by c hat plus 1 multiplied by b hat let's get b hat b hat is just b divided by the norm of b the norm of b is 5 then it's 1 divided by 5 multiplied by 4i plus let's get a a equals 2c 2 divided by 5 multiplied by 3 i plus 4 j plus 1 multiplied by b 1 divided by 5 multiplied by 4i plus 3 j doing the sums it equals minus 6 divided by 5 plus 4 divided by 5 gives minus 2 divided by 5 i 4 multiplied by 2 divided by 5 plus 3 divided by 5 gives 8 divided by 5 j sorry this is not 8 it it's 2 multiplied by 4 8 plus 3 equals 11, 11 divided by 5.

04:59 This is one option for a.

05:02 Another option is to consider that c in the opposite direction.

05:08 Then we have a equals minus 2 divided by 5 minus 3i plus 4j, plus 1 divided by 5, 4i plus 3g...

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