Solveeit Logo
Login

Question

Mathematics Question on Vector Algebra

Let three vectors a⃗=αi^+4j^+2k^\vec{a} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k},
b⃗=5i^+3j^+4k^\vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k},
c⃗=xi^+yj^+zk^\vec{c} = x \hat{i} + y \hat{j} + z \hat{k} from a triangle such that c⃗=a⃗−b⃗\vec{c} = \vec{a} - \vec{b} and the area of the triangle is 565 \sqrt{6}. If α\alpha is a positive real number, then ∣c⃗∣2|\vec{c}|^2 is:

A

16

B

14

C

12

D

10

Answer

14

Explanation

Solution

Step 1: Expression for c⃗\vec{c}

The vector c⃗\vec{c} is given as:

c⃗=a⃗−b⃗\vec{c} = \vec{a} - \vec{b}.

Substitute a⃗=αi^+4j^+2k^\vec{a} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k} and b⃗=5i^+3j^+4k^\vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}:

c⃗=(α−5)i^+(4−3)j^+(2−4)k^\vec{c} = (\alpha - 5)\hat{i} + (4 - 3)\hat{j} + (2 - 4)\hat{k}.

Thus:

c⃗=(α−5)i^+j^−2k^\vec{c} = (\alpha - 5)\hat{i} + \hat{j} - 2\hat{k}.

Step 2: Area of the triangle

The area of the triangle is given as:

Area=12∣a⃗×c⃗∣.\text{Area} = \frac{1}{2} |\vec{a} \times \vec{c}|.

Substitute Area=56\text{Area} = 5\sqrt{6}:

12∣a⃗×c⃗∣=56.\frac{1}{2} |\vec{a} \times \vec{c}| = 5\sqrt{6}.

∣a⃗×c⃗∣=106.|\vec{a} \times \vec{c}| = 10\sqrt{6}.

Step 3: Cross product a⃗×c⃗\vec{a} \times \vec{c}

The cross product a⃗×c⃗\vec{a} \times \vec{c} is given by:

a⃗×c⃗=∣i^j^k^ α42 α−51−2∣.\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ \alpha & 4 & 2 \\\ \alpha - 5 & 1 & -2 \end{vmatrix}.

Expanding the determinant:

a⃗×c⃗=i^∣42 1−2∣−j^∣α2 α−5−2∣+k^∣α4 α−51∣.\vec{a} \times \vec{c} = \hat{i} \begin{vmatrix} 4 & 2 \\\ 1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} \alpha & 2 \\\ \alpha - 5 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} \alpha & 4 \\\ \alpha - 5 & 1 \end{vmatrix}.

Calculate each minor:

1. For i^\hat{i}: ∣42 1−2∣=(4)(−2)−(2)(1)=−8−2=−10.\begin{vmatrix} 4 & 2 \\\ 1 & -2 \end{vmatrix} = (4)(-2) - (2)(1) = -8 - 2 = -10.

2. For j^\hat{j}: ∣α2 α−5−2∣=(α)(−2)−(α−5)(2)=−2α−2α+10=−4α+10.\begin{vmatrix} \alpha & 2 \\\ \alpha - 5 & -2 \end{vmatrix} = (\alpha)(-2) - (\alpha - 5)(2) = -2\alpha - 2\alpha + 10 = -4\alpha + 10.

3. For k^\hat{k}: ∣α4 α−51∣=(α)(1)−(α−5)(4)=α−4α+20=−3α+20.\begin{vmatrix} \alpha & 4 \\\ \alpha - 5 & 1 \end{vmatrix} = (\alpha)(1) - (\alpha - 5)(4) = \alpha - 4\alpha + 20 = -3\alpha + 20.

Thus:

a⃗×c⃗=−10i^−(−4α+10)j^+(−3α+20)k^.\vec{a} \times \vec{c} = -10\hat{i} - (-4\alpha + 10)\hat{j} + (-3\alpha + 20)\hat{k}.

a⃗×c⃗=−10i^+(4α−10)j^+(−3α+20)k^.\vec{a} \times \vec{c} = -10\hat{i} + (4\alpha - 10)\hat{j} + (-3\alpha + 20)\hat{k}.

Step 4: Magnitude of a⃗×c⃗\vec{a} \times \vec{c}

The magnitude is:

∣a⃗×c⃗∣=(−10)2+(4α−10)2+(−3α+20)2.|\vec{a} \times \vec{c}| = \sqrt{(-10)^2 + (4\alpha - 10)^2 + (-3\alpha + 20)^2}.

∣a⃗×c⃗∣=100+(16α2−80α+100)+(9α2−120α+400).|\vec{a} \times \vec{c}| = \sqrt{100 + (16\alpha^2 - 80\alpha + 100) + (9\alpha^2 - 120\alpha + 400)}.

∣a⃗×c⃗∣=25α2−200α+600.|\vec{a} \times \vec{c}| = \sqrt{25\alpha^2 - 200\alpha + 600}.

Set ∣a⃗×c⃗∣=106|\vec{a} \times \vec{c}| = 10\sqrt{6}:

25α2−200α+600=106.\sqrt{25\alpha^2 - 200\alpha + 600} = 10\sqrt{6}.

Square both sides:

25α2−200α+600=600.25\alpha^2 - 200\alpha + 600 = 600.

25α(α−8)=0.25\alpha(\alpha - 8) = 0.

Since α>0\alpha > 0, α=8.\alpha = 8.

Step 5: Calculate ∣c⃗∣2|\vec{c}|^2

Substitute α=8\alpha = 8 into c⃗\vec{c}:

c⃗=3i^+j^−2k^.\vec{c} = 3\hat{i} + \hat{j} - 2\hat{k}.

The magnitude squared is:

∣c⃗∣2=32+(1)2+(−2)2.|\vec{c}|^2 = 3^2 + (1)^2 + (-2)^2.

∣c⃗∣2=9+1+4=14.|\vec{c}|^2 = 9 + 1 + 4 = 14.

Final Answer: Option (2).