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Question: Find the value of \(\lambda \) for which the angle between the vectors \[\vec a = 2{\lambda ^2}\hat ...

Find the value of λ\lambda for which the angle between the vectors a⃗=2λ2i^+4λj^+k^\vec a = 2{\lambda ^2}\hat i + 4\lambda \hat j + \hat k and b⃗=7i^−2j^+λk^\vec b = 7\hat i - 2\hat j + \lambda \hat k is obtuse.

Explanation

Solution

In this particular sum, there are 22 important things which the student has to keep in mind, firstly that whenever a sum related to angles is given he/she should always use dot product instead of the cross product. Secondly, since the angle is abuse that is the value of θ\theta would be greater than 90∘{90^ \circ }, the value of cos⁡θ\cos \theta would be negative, thus the quadratic equation would have an inequality i.e. <0 < 0. After solving the student will get a range of answers for which the angle is obtuse.

Complete step by step solution:
Since we have that the angle between the vectors is obtuse, we will be using the dot product.
a⃗.b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a.\vec b = |\vec a||\vec b|\cos \theta
∴cos⁡θ=a⃗.b⃗∣a⃗∣∣b⃗∣\therefore \cos \theta = \dfrac{{\vec a.\vec b}}{{|\vec a||\vec b|}}
Also we want the angle between 22vectors to be obtuse, cos⁡θ<0\cos \theta < 0,
∴a⃗.b⃗∣a⃗∣∣b⃗∣<0\therefore \dfrac{{\vec a.\vec b}}{{|\vec a||\vec b|}} < 0
(2λ2i^+4λj^+k^).(7i^−2j^+λk^)<0........(1)(2{\lambda ^2}\hat i + 4\lambda \hat j + \hat k).(7\hat i - 2\hat j + \lambda \hat k) < 0........(1)
We have ignored the denominator, as we have 00 on the RHS.
Simplifying Equation 11,
14λ2−8λ+λ<0........(2)14{\lambda ^2} - 8\lambda + \lambda < 0........(2)
7λ(2λ−1)<0........(3)7\lambda (2\lambda - 1) < 0........(3)
0<λ<120 < \lambda < \dfrac{1}{2}

Thus we can say that the angle between the vectors is obtuse if the value of λ\lambda is in the range of 0<λ<120 < \lambda < \dfrac{1}{2} or λ∈(0,12)\lambda \in (0,\dfrac{1}{2}).

Note: Students should always use dot products whenever they have to find a vector or the angle between the vectors. This is because it is less complicated compared to cross-product which consists of determinants to find the value. Whenever a student is given a sum that is from the chapter inequalities, he should first equate it to 00 and then after obtaining the answer should replace the == with the required sign. Also, the student should keep in mind that whenever a problem on a quadratic equation is asked, the student should not strike off the common terms, instead, they should take common as we did in the above sum. If he strikes off the term the student will get only one root which would make the answer wrong.