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Question: Find the value of \(\lambda \) so that the vector \(A = 2\widehat i + \lambda \widehat j - \widehat ...

Find the value of λ\lambda so that the vector A=2i^+λj^k^,B=4i^2j^2k^A = 2\widehat i + \lambda \widehat j - \widehat k,B = 4\widehat i - 2\widehat j - 2\widehat k are perpendicular to each other

Explanation

Solution

A vector in the physics is defined as a quantity that has both magnitude and direction. It is generally represented by an arrow whose direction is the same as that of the quantity and the length is proportional to the quantity's magnitude. Although a vector has both magnitude and the direction, it does not have a position. That is, as long as the vector is not altered if it is displaced parallel to itself its length is not changed

The other rules of a vector manipulation are subtraction and multiplication by a scalar, scalar multiplication is also known as the dot product or inner product, vector multiplication (it is also known as the cross product), and differentiation. There is no process that corresponds to the dividing by a vector.

Complete step by step solution:
To find the value of λ\lambda we have to multiply the both vector so that we get the value of λ\lambda
A.B=(2×4)(λ×(2))+(1×(2))\overrightarrow A .\overrightarrow B = (2 \times 4) - (\lambda \times ( - 2)) + ( - 1 \times ( - 2))
After simplifying the above equation we get
8+2λ+2=08 + 2\lambda + 2 = 0
So now we want the value of λ\lambda so take it outside after taking it outside we get the value that is
10=2λ\Rightarrow 10 = 2\lambda
Now divide 1010by 22after calculating this we get the value of λ\lambda
Therefore λ=5\lambda = 5

Hence we got the value of λ\lambda

Note:
The vector is said to be perpendicular if you draw them perpendicular, If the angle between them is 9090^\circ , If a dot product of them is zero

If a vector product of them is maximum, If on the projection of the vectors one relation remains the same with the triangle formed by using the length of vectors If the vectors are along with any two of the coordinate axes.