Q.15. Let and . Then the square of the area of the triangle with adjacent sides determined by the v

Q: What is the correct answer for this exam Mathematics Q15?

The correct answer is C. Given vectors $\vec{a}=2\hat{i}+3\hat{j}+3\hat{k}$ and $\vec{b}=6\hat{i}+3\hat{j}+3\hat{k}$.Let $\vec{u} = 2\vec{a}+3\vec{b}$ and $\vec{v} = \vec{a}-\vec{b}$.$\vec{u} = 2(2\hat{i}+3\hat{j}+3\hat{k}) + 3(6\hat{i}+3\hat{j}+3\hat{k})$$= (4\hat{i}+6\hat{j}+6\hat{k}) + (18\hat{i}+9\hat{j}+9\hat{k})$$= (4+18)\hat{i} + (6+9)\hat{j} + (6+9)\hat{k} = 22\hat{i}+15\hat{j}+15\hat{k}$.$\vec{v} = (2\hat{i}+3\ha

Q: Q.15. Let and . Then the square of the area of the triangle with adjacent sides determined by the vectors and is:

The correct answer is C. Given vectors $\vec{a}=2\hat{i}+3\hat{j}+3\hat{k}$ and $\vec{b}=6\hat{i}+3\hat{j}+3\hat{k}$.Let $\vec{u} = 2\vec{a}+3\vec{b}$ and $\vec{v} = \vec{a}-\vec{b}$.$\vec{u} = 2(2\hat{i}+3\hat{j}+3\hat{k}) + 3(6\hat{i}+3\hat{j}+3\hat{k})$$= (4\hat{i}+6\hat{j}+6\hat{k}) + (18\hat{i}+9\hat{j}+9\hat{k})$$= (4+18)\hat{i} + (6+9)\hat{j} + (6+9)\hat{k} = 22\hat{i}+15\hat{j}+15\hat{k}$.$\vec{v} = (2\hat{i}+3\ha

Q: What are the options for this this exam Mathematics question?

A. 450 · B. 900 · C. 1800 · D. 2400

Question

Q.15. Let  and . Then the square of the area of the triangle with adjacent sides determined by the vectors  and  is:

FirstInTest · Accepted Answer

Given vectors $\vec{a}=2\hat{i}+3\hat{j}+3\hat{k}$ and $\vec{b}=6\hat{i}+3\hat{j}+3\hat{k}$.Let $\vec{u} = 2\vec{a}+3\vec{b}$ and $\vec{v} = \vec{a}-\vec{b}$.$\vec{u} = 2(2\hat{i}+3\hat{j}+3\hat{k}) + 3(6\hat{i}+3\hat{j}+3\hat{k})$$= (4\hat{i}+6\hat{j}+6\hat{k}) + (18\hat{i}+9\hat{j}+9\hat{k})$$= (4+18)\hat{i} + (6+9)\hat{j} + (6+9)\hat{k} = 22\hat{i}+15\hat{j}+15\hat{k}$.$\vec{v} = (2\hat{i}+3\hat{j}+3\hat{k}) - (6\hat{i}+3\hat{j}+3\hat{k})$$= (2-6)\hat{i} + (3-3)\hat{j} + (3-3)\hat{k} = -4\hat{i}+0\hat{j}+0\hat{k} = -4\hat{i}$.The area of the triangle with adjacent sides $\vec{u}$ and $\vec{v}$ is $\frac{1}{2}|\vec{u} 	imes \vec{v}|$.$\vec{u} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 22 & 15 & 15 \ -4 & 0 & 0 \end{vmatrix}$$= \hat{i}(15 	imes 0 - 15 	imes 0) - \hat{j}(22 	imes 0 - 15 	imes -4) + \hat{k}(22 	imes 0 - 15 	imes -4)$$= \hat{i}(0) - \hat{j}(60) + \hat{k}(60) = -60\hat{j}+60\hat{k}$.$|\vec{u} 	imes \vec{v}| = \sqrt{(-60)^2 + (60)^2} = \sqrt{3600+3600} = \sqrt{7200} = \sqrt{3600 	imes 2} = 60\sqrt{2}$.Area of triangle $= \frac{1}{2} |\vec{u} 	imes \vec{v}| = \frac{1}{2} (60\sqrt{2}) = 30\sqrt{2}$.The square of the area of the triangle is $(30\sqrt{2})^2 = 900 	imes 2 = 1800$.

Answer

450

Answer

900

Answer

1800

Answer

2400

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