Search
You can find the results of your search below.
Fulltext results:
- Question 12 & 13, Exercise 3.3
- are the midpoints of sides shown\\ \begin{align}\therefore \quad \overrightarrow{O D}&=\dfrac{\vec{b}+\v... htarrow{O D} \perp \overrightarrow{B C} \quad \\ \therefore \quad \overrightarrow{O D} \cdot \overrightar... rightarrow{O E} \perp \overrightarrow{C A} \quad \therefore \quad \overrightarrow{O E} \cdot \overrightar
- Question 7 Exercise 3.5
- ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \vec{v} \times \ve... ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \vec{v} \times \ve... olution==== Since the given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \bar{v} \times \ve
- Question 1 Exercise 3.4
- ec{a}&=2 \hat{i}-3 \hat{j}\\ \vec{b}&=\hat{k} \\ \therefore \vec{a} \times \vec{b}&=\left|\begin{array}{c... {k}\\ \vec{b}&=6 \hat{i}+2 \hat{j}-3 \hat{k} \\ \therefore \vec{a} \times \vec{b}&=\left|\begin{array}{c
- Question 5 Exercise 3.4
- htarrow{P Q} \times \overrightarrow{P R}|&=30 \\ \therefore \text { Area of triangle }& =\dfrac{1}{2}|\ov... {P Q} \times \overrightarrow{P R}|&=\sqrt{76} \\ \therefore \text { Area of triangle }&=\dfrac{1}{2} | \v
- Question 4 and 5 Exercise 3.3
- at is orthogonal to both $\vec{a}$ and $\vec{b}$. Therefore, \begin{align}\vec{c}&=\vec{a} \times \vec{b}
- Question 9 & 10, Exercise 3.3
- )$$ \begin{align}\rightarrow W&=2.7+3 .-1+1.1 \\ \therefore W&=14-3+1=12 \text { units. }\end{align}
- Question 11, Exercise 3.3
- w \vec{a} \cdot \vec{c}&=6 - 2-4=0 .\end{align} $\therefore \quad \vec{a} \cdot \vec{c}$. or sides repres
- Question 9 Exercise 3.4
- nals, so $E$ is the midpoint of both diagonals.\\ Therefore, \\ \begin{align}\overrightarrow{A E}&=\overr
- Question 8 Exercise 3.5
- at{j}-5 \hat{k}-(2 \hat{i}+3 \hat{j}+\hat{k}) \\ \therefore \vec{b}&=2 \hat{i}-\hat{j} - 6 \hat{k} \\ \ve
- Question 6 & 7 Review Exercise 3
- olution==== Since the given vectors are coplanar, therefore, \begin{align}\vec{a} \cdot \vec{b} \times \
- Question 8 & 9 Review Exercise 3
- \ -1 & 3 & 0 \\ 2 & -3 & 2 \end{array}\right| \\ \therefore \vec{a} \times \vec{b} &=(\hat{i}+2 \hat{j} \