a) Find the equation of the ellipse whose major axis is twice its minor axis and passes through the point (0, 1) . [3] b) The position vectors of the vertices of are 7 j + 10 k , i + 6 j + 6 k and 4…

a) Equation of the Ellipse

Given:

• Major axis is twice the minor axis.
• Ellipse passes through the point $(0,1)$.

Step 1: Standard Form of Ellipse The standard form of an ellipse centered at the origin with major axis along the x-axis is: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ where:

• $a$ is the semi-major axis,
• $b$ is the semi-minor axis.

Step 2: Relationship Between Axes Given that the major axis is twice the minor axis: $$2a=2\times 2b⟹a=2b$$

Step 3: Substitute into the Ellipse Equation $$\frac{x^2}{(2b{)}^2}+\frac{y^2}{b^2}=1⟹\frac{x^2}{4b^2}+\frac{y^2}{b^2}=1$$

Step 4: Use the Given Point $(0,1)$ Substitute $x=0$ and $y=1$ into the equation: $$\frac{0^2}{4b^2}+\frac{1^2}{b^2}=1⟹\frac{1}{b^2}=1⟹b^2=1⟹b=1$$

Step 5: Find $a$ Since $a=2b$: $$a=2\times 1=2⟹a^2=4$$

Step 6: Final Equation of the Ellipse $$\frac{x^2}{4}+y^2=1$$

Answer: $$\boxed{\frac{x^2}{4}+y^2=1}$$

---

b) Proving $\Delta ABC$ is an Isosceles Right-Angled Triangle

Given:

• Position vectors of vertices:
  • $A=7\vec{j}+10\vec{k}=(0,7,10)$,
  • $B=-\vec{i}+6\vec{j}+6\vec{k}=(-1,6,6)$,
  • $C=-4\vec{i}+9\vec{j}+6\vec{k}=(-4,9,6)$.

Step 1: Find Vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{BC}$ $$\overrightarrow{AB}=B-A=(-1-0,6-7,6-10)=(-1,-1,-4)$$ $$\overrightarrow{AC}=C-A=(-4-0,9-7,6-10)=(-4,2,-4)$$ $$\overrightarrow{BC}=C-B=(-4-(-1),9-6,6-6)=(-3,3,0)$$

Step 2: Compute Magnitudes of Vectors $$|\overrightarrow{AB}|=\sqrt{(-1{)}^2+(-1{)}^2+(-4{)}^2}=\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}$$ $$|\overrightarrow{AC}|=\sqrt{(-4{)}^2+2^2+(-4{)}^2}=\sqrt{16+4+16}=\sqrt{36}=6$$ $$|\overrightarrow{BC}|=\sqrt{(-3{)}^2+3^2+0^2}=\sqrt{9+9+0}=\sqrt{18}=3\sqrt{2}$$

Step 3: Check for Isosceles Property Since $|\overrightarrow{AB}|=|\overrightarrow{BC}|=3\sqrt{2}$, $\Delta ABC$ is isosceles with $AB=BC$.

Step 4: Check for Right Angle Compute the dot product of $\overrightarrow{AB}$ and $\overrightarrow{BC}$: $$\overrightarrow{AB}\cdot \overrightarrow{BC}=(-1)(-3)+(-1)(3)+(-4)(0)=3-3+0=0$$ Since the dot product is zero, $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are perpendicular. Thus, $\angle B$ is $90^{\circ }$.

Conclusion: $\Delta ABC$ is an isosceles right-angled triangle with the right angle at $B$.

---

c) Proving $a^2+b^2+c^2-2(bc\cos A+ca\cos B+ab\cos C)=0$

Given: In any triangle $ABC$, we need to prove: $$a^2+b^2+c^2-2(bc\cos A+ca\cos B+ab\cos C)=0$$

Step 1: Recall the Cosine Rule For any triangle: $$\cos A=\frac{b^2+c^2-a^2}{2bc},\cos B=\frac{a^2+c^2-b^2}{2ac},\cos C=\frac{a^2+b^2-c^2}{2ab}$$

Step 2: Substitute $\cos A$, $\cos B$, and $\cos C$ into the Expression $$bc\cos A=bc\cdot \frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-a^2}{2}$$ $$ca\cos B=ca\cdot \frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-b^2}{2}$$ $$ab\cos C=ab\cdot \frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2-c^2}{2}$$

Step 3: Compute $2(bc\cos A+ca\cos B+ab\cos C)$ $$2(bc\cos A+ca\cos B+ab\cos C)=2\left(\frac{b^2+c^2-a^2}{2}+\frac{a^2+c^2-b^2}{2}+\frac{a^2+b^2-c^2}{2}\right)$$ $$=(b^2+c^2-a^2)+(a^2+c^2-b^2)+(a^2+b^2-c^2)$$ $$=a^2+b^2+c^2$$

Step 4: Substitute Back into the Original Expression $$a^2+b^2+c^2-2(bc\cos A+ca\cos B+ab\cos C)=a^2+b^2+c^2-(a^2+b^2+c^2)=0$$

Conclusion: The given identity is proven: $$\boxed{a^2+b^2+c^2-2(bc\cos A+ca\cos B+ab\cos C)=0}$$