a) The scalar product of two vectors and cross product of two vectors are interrelated. Explain. [3] b) If the cosines of two angles of a triangle are proportional to the opposite sides, show that…

a) Relationship between Scalar Product and Cross Product of Two Vectors

The scalar product (dot product) and cross product (vector product) of two vectors are mathematically related through the magnitude of the vectors and the angle between them. Their relationship is derived from the orthogonality of the three-dimensional coordinate system.

Key Relationships:

1. Magnitude of Cross Product: For two vectors $\mathbf{u}=(u_1,u_2,u_3)$ and $\mathbf{v}=(v_1,v_2,v_3)$, the magnitude of their cross product is given by: $$|\mathbf{u}\times \mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin \theta$$ where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$.

2. Scalar Product: The scalar product is defined as: $$\mathbf{u}\cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos \theta$$

3. Combining Both: Using the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$, we can express $\sin \theta$ in terms of $\cos \theta$: $$\sin \theta =\sqrt{1-{\cos }^2\theta }$$ Substituting into the magnitude of the cross product: $$|\mathbf{u}\times \mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sqrt{1-{\cos }^2\theta }$$

4. Explicit Relationship: The scalar triple product (a scalar formed by the dot product of one vector with the cross product of the other two) is zero if the three vectors are coplanar: $$\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})=0$$ This shows that the cross product is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$, while the scalar product projects one vector onto another.

Geometric Interpretation:

• The scalar product measures the projection of one vector onto another.
• The cross product measures the area of the parallelogram formed by the two vectors, which is maximized when $\theta =90^{\circ }$ (i.e., $\sin \theta =1$).

---

b) Proving a Triangle is Isosceles Given Proportional Cosines

Given: In a triangle $ABC$, the cosines of angles $A$ and $B$ are proportional to the lengths of the opposite sides $a$ and $b$, respectively: $$\frac{\cos A}{a}=\frac{\cos B}{b}=k\text{(some constant)}$$

To Show: The triangle is isosceles (i.e., $a=b$ or $A=B$).

Proof:

1. By the Law of Cosines, we have: $$\cos A=\frac{b^2+c^2-a^2}{2bc},\cos B=\frac{a^2+c^2-b^2}{2ac}$$

2. Given $\frac{\cos A}{a}=\frac{\cos B}{b}$, substitute the expressions: $$\frac{\frac{b^2+c^2-a^2}{2bc}}{a}=\frac{\frac{a^2+c^2-b^2}{2ac}}{b}$$

3. Simplify both sides: $$\frac{b^2+c^2-a^2}{2abc}=\frac{a^2+c^2-b^2}{2abc}$$

4. Multiply both sides by $2abc$ (since $a,b,c>0$): $$b^2+c^2-a^2=a^2+c^2-b^2$$

5. Cancel $c^2$ and rearrange: $$b^2-a^2=a^2-b^2⟹2b^2=2a^2⟹b^2=a^2⟹b=a$$

Conclusion: Since $a=b$, the triangle is isosceles with $A=B$.

---

c) Condition for the Line to be Normal to the Parabola

Given: The parabola $x^2=4ay$ and the line $ax+by+c=0$.

To Find: The condition under which the line is normal to the parabola.

Solution:

1. Equation of the Parabola: The standard parabola $x^2=4ay$ has its focus at $(0,a)$ and directrix $y=-a$.

2. Slope of the Tangent to the Parabola: Differentiate implicitly: $$2x=4a\frac{dy}{dx}⟹\frac{dy}{dx}=\frac{x}{2a}$$ The slope of the tangent at any point $(x_1,y_1)$ is $m_{\text{tangent}}=\frac{x_1}{2a}$.

3. Slope of the Normal: The normal is perpendicular to the tangent, so its slope is: $$m_{\text{normal}}=-\frac{2a}{x_1}$$

4. Slope of the Given Line: Rewrite the line $ax+by+c=0$ in slope-intercept form: $$y=-\frac{a}{b}x-\frac{c}{b}$$ The slope of the line is $m_{\text{line}}=-\frac{a}{b}$.

5. Equating Slopes: For the line to be normal to the parabola, its slope must equal the slope of the normal: $$-\frac{a}{b}=-\frac{2a}{x_1}⟹\frac{a}{b}=\frac{2a}{x_1}$$ Simplify (assuming $a\ne 0$): $$\frac{1}{b}=\frac{2}{x_1}⟹x_1=2b$$

6. Point of Contact: Since $(x_1,y_1)$ lies on the parabola: $$x_1^2=4a{y}_1⟹(2b{)}^2=4a{y}_1⟹y_1=\frac{4b^2}{4a}=\frac{b^2}{a}$$

7. Substitute into the Line Equation: The line passes through $(2b,\frac{b^2}{a})$: $$a(2b)+b\left(\frac{b^2}{a}\right)+c=0⟹2ab+\frac{b^3}{a}+c=0$$ Multiply by $a$ to eliminate the denominator: $$2a^{2}b+b^3+ac=0$$

Final Condition: The line $ax+by+c=0$ is normal to the parabola $x^2=4ay$ if and only if: $$\boxed{2a^{2}b+b^3+ac=0}$$