Solveeit Logo
Login

Question

Mathematics Question on Vectors

Let
a⃗=αi^+j^−k^ and b⃗=2i^+j^−αk^,α>0\vec{a}=α\hat{i}+\hat{j}−\hat{k}\ and\ \vec{b}=2\hat{i}+\hat{j}−α\hat{k},α>0.
If the projection of a⃗×b⃗\vec{a}×\vec{b} on the vector −i^+2j^−2k^−\hat{i}+2\hat{j}−2\hat{k}
is 30, then α is equal to

A

152\frac{15}{2}

B

8

C

132\frac{13}{2}

D

7

Answer

7

Explanation

Solution

Given: a⃗=(α,1,−1)\vec{a}=(α,1,−1)
and
b⃗=(2,1,−α)\vec{b}=(2,1,−α)
\vec{c}=\vec{a}×\vec{b}=$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\\ \alpha & 1 & -1 \\\ 2 & 1 &-\alpha \end{vmatrix}
=(−α+1)i^+(α2−2)j^+(α−2)k^=(−α+1)\hat{i}+(α2−2)\hat{j}+(α−2)\hat{k}
Projection of c⃗\vec{c} on d⃗=−i^+2j^−2k^\vec{d}=−\hat{i}+2\hat{j}−2\hat{k}
=|\vec{c}⋅\frac{\vec{d}}{|d|}|=30 \left\\{ Given\right\\}
⇒=∣α−1−4+2α2−2α+41+4+4∣=30⇒ =|\frac{α−1−4+2α^2−2α+4}{\sqrt{1+4+4}}|=30
On solving
α=−132α=\frac{−13}{2}
(Rejected as α> 0)
and α = 7
So, the correct option is (D): 7