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Easy

Question

The vectors a with not stretchy bar on top comma b with not stretchy bar on top comma c with not stretchy bar on top are equal in length and pair wise make equal angles. If a with not stretchy bar on top equals l with not stretchy bar on top plus ȷ with not stretchy bar on top comma b with not stretchy bar on top equals ȷ with not stretchy bar on top plus k with not stretchy bar on top, then c with not stretchy bar on top equals

  1. left parenthesis 1 comma 0 comma 1 right parenthesis
  2. left parenthesis 1 comma 1 comma 1 right parenthesis
  3. left parenthesis 0 comma 1 comma 1 right parenthesis
  4. left parenthesis 1 comma 1 comma 0 right parenthesis

hintHint:

We are given three vectors of same length. The angle between the pairs of the vector is same too. We are given the two vectors. We have to find the remaining vector. We will assume the variables for the remaining vector. We will use it for all the given conditions. Then, we will find the values of the variable hence the vector.

The correct answer is: left parenthesis 1 comma 0 comma 1 right parenthesis


    The given vectors are a with rightwards arrow on top comma b with rightwards arrow on top a n d space c with rightwards arrow on top.
    a with rightwards arrow on top equals i with hat on top plus j with hat on top space
b with rightwards arrow on top equals j with hat on top plus k with hat on top
    L e t space c with space rightwards arrow on top equals x i with hat on top plus y j with hat on top space plus z k with hat on top
    All the three vectors are of equal lengths.
    open vertical bar a with rightwards arrow on top open vertical bar equals close vertical bar b with rightwards arrow on top open vertical bar equals close vertical bar c with rightwards arrow on top close vertical bar        ...(1)
    open vertical bar a with rightwards arrow on top close vertical bar equals square root of 1 plus 1 end root
space space space space space space equals square root of 2
    So, the length of all the vectors will be √2
    Let the angle between each pair from the given vectors be A.
    Let's write the dot product of each pair.
    a with rightwards arrow on top. b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos A
    open parentheses i with hat on top plus j with hat on top close parentheses. open parentheses j with hat on top plus k with hat on top close parentheses equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
space 0 space plus space 1 space plus space 0 space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
R e a r r a n g e
open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space equals space 1 space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    The next pair will be
    b with rightwards arrow on top. c with rightwards arrow on top equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos A
    open parentheses j with hat on top plus k with hat on top close parentheses times open parentheses x i with hat on top plus y j with hat on top plus z k with hat on top close parentheses equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos A
y space plus space z space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space space space space space space space space space space space space space space... left parenthesis space u sin g space 1 right parenthesis
y space plus space z space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis u sin g space 2 right parenthesis space space space space space space space space space space

    c with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
    open parentheses x i with hat on top plus y j with hat on top plus z k with hat on top close parentheses. open parentheses i with hat on top plus j with hat on top close parentheses equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A
x space plus space y space plus space 0 space equals space open vertical bar a with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos A space space space space space space space... left parenthesis u sin g space 1 right parenthesis
x space plus space y space equals space 1 space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis u sin g space 2 right parenthesis
    Now we will solve the equation
    y + z = 1       ...(3)
    x + y = 1      ...(4)
    Subtracting  (4) from (3)
    z - x = 0
    x = z
    Now, we know that length of all vectors is √2
    open vertical bar c with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root
square root of 2 space equals square root of x squared plus y squared plus space z squared end root
space S q u a r e space b o t h space t h e space s i d e s space a n d space r e a r r a n g e
x squared space plus space y squared space plus space z squared space equals space 2
w e space k n o w space x space equals space z
2 z squared space plus space y squared space equals space 2
    We will rearrange equation (3) and substitute in the above equation.
    y + z = 1
    y = 1 - z
    2 z squared plus y squared equals 2
2 z squared plus left parenthesis 1 space minus space z right parenthesis squared space equals space 2 2 z squared space plus space 1 space minus space 2 z space plus space z squared space equals space 2
3 z squared space minus 2 z space minus 1 space equals space 0
F a c t o r i s e
3 z squared space minus 3 z space plus space z space minus space 1 space equals space 0
3 z left parenthesis z space minus space 1 right parenthesis space plus space 1 left parenthesis z space minus space 1 right parenthesis space equals space 0
left parenthesis 3 z space plus space 1 right parenthesis left parenthesis z space minus space 1 right parenthesis space equals space 0
z space equals space minus 1 third space o r space z space equals space 1
    If we see the options, we don't have the value of z component as fraction. So, we will just focus on the value of z = 1
    y = 1 - z
    y = 1 - 1
    y = 0
    x + y = 1
    x + 0 = 1
    x = 1
    c with rightwards arrow on top equals i with hat on top plus 0 j with overparenthesis on top plus k with hat on top
    The solution is (1,0,1).

    For such questions, concept of scalar product is important. When we take a dot product of two vectors, we get a scalar value. This is called as scalar product.

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