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Question: For what value of \(\lambda \) , the vectors \(\overrightarrow{a}=2\overrightarrow{i}+\lambda \vec{j...

For what value of λ\lambda , the vectors a=2i+λj+k\overrightarrow{a}=2\overrightarrow{i}+\lambda \vec{j}+\vec{k} and b=i+2j3k\vec{b}=-\vec{i}+2\vec{j}-3\vec{k} are orthogonal to each other?

Explanation

Solution

Hint : Here we have to find for what value of λ\lambda the given two vectors are orthogonal to each other. If two vectors are orthogonal, then their dot product is zero. So firstly we will form an equation where we will put the dot product of the two given vectors equal to 00 . Then we will use the scalar product results and solve the obtained equation to get our desired answer.

Complete step-by-step answer :
We have to find the value of λ\lambda using the condition that vector a=2i+λj+k\overrightarrow{a}=2\overrightarrow{i}+\lambda \vec{j}+\vec{k} and b=i+2j3k\vec{b}=-\vec{i}+2\vec{j}-3\vec{k} are orthogonal to each other.
Firstly as we know that if dot products of two vectors are 00 then they are said to be orthogonal to each other.
So we will from an equation such that the dot product of the two vectors given is put equal to 00 as follows:
a.b=0\vec{a}.\vec{b}=0
On substituting the value of the two vectors above we get,
(2i+λj+k).(i+2j3k)=0\left( 2\overrightarrow{i}+\lambda \vec{j}+\vec{k} \right).\left( -\vec{i}+2\vec{j}-3\vec{k} \right)=0 …….(1)\left( 1 \right)
Now as we know that the result of dot/scalar product is as follows:
i.i=1\vec{i}.\vec{i}=1 j.j=1\vec{j}.\vec{j}=1 k.k=1\vec{k}.\vec{k}=1 i.j=0\vec{i}.\vec{j}=0 i.k=0\vec{i}.\vec{k}=0 j.k=0\vec{j}.\vec{k}=0
As we can see from above result that all the dot product of two different unit vectors is zero and same unit vector is one we can rewrite the equation (1) as follows:
2×1+λ×2+1×3=02\times -1+\lambda \times 2+1\times -3=0
2+2λ3=0-2+2\lambda -3=0
Taking unknown variable one side and rest value on another side we get,
2λ=3+22\lambda =3+2
λ=52\Rightarrow \lambda =\dfrac{5}{2}
So we got the value of λ\lambda as 52\dfrac{5}{2} or in decimal 2.52.5 .
Hence for λ=2.5\lambda =2.5 the vectors a=2i+λj+k\overrightarrow{a}=2\overrightarrow{i}+\lambda \vec{j}+\vec{k} and b=i+2j3k\vec{b}=-\vec{i}+2\vec{j}-3\vec{k} are orthogonal to each other.

Note : Vectors quantity or vectors are those quantities which have magnitude as well as direction. A vector whose magnitude is proportional to the length ABAB and whose direction is from AA to BB is usually denoted by AB\overrightarrow{AB} . There are many types of vectors such as Like Vectors, Unlike Vectors, Null Vectors and Equal Vectors etc. Unit vectors are those vectors whose magnitude is unity and we denote unit vectors along the OX,OY,OZOX,OY,OZ axis by i.j,k\vec{i}.\vec{j},\vec{k} respectively.