Long Question

Prove by vector method that the circumference angle ACB of a semi-circle with a diameter AB is a right-angle.

Given: O is the origin. ACB is a semi-circle. AB is diameter of semi-circle. To prove: ∠ACB = 90°
Construction: Join O to C.
Proof:
Statement Reason
i) $\overrightarrow{BC}$ = $\overrightarrow{BO} + \overrightarrow{OC}$ = $\overrightarrow{OC} - \overrightarrow{OB}$ i) Triangle law of vector addition.
ii) $\overrightarrow{CA}$ = $\overrightarrow{CO} + \overrightarrow{OA}$
= $\overrightarrow{OA} - \overrightarrow{OC}$
ii) Triangle law of vector addition.
iii) $\overrightarrow{BC}$ = $\overrightarrow{OC} - (-\overrightarrow{OA})$ = $\overrightarrow{OC} + \overrightarrow{OA}$ iii) $\overrightarrow{OB} = -\overrightarrow{OA}$
iv) $\overrightarrow{BC}$.$\overrightarrow{CA}$ = $(\overrightarrow{OC} + \overrightarrow{OA}).(\overrightarrow{OC} - \overrightarrow{OA})$ = $(\overrightarrow{OC})^2 - (\overrightarrow{OA})^2$ iv) Multiplying statement (ii) and (iii)
v) $\overrightarrow{BC}$.$\overrightarrow{CA}$ = $(\overrightarrow{OC})^2 - (\overrightarrow{OC})^2$ = 0 v) OA = OC (radius)
vi) ∴ ∠BCA = 90° vi) Dot product of $\overrightarrow{BC}$ and $\overrightarrow{CA}$ is zero.
Hence, It is proved that ∠ACB is a right-angle.

PQ is a diameter of a circle with the centre X and M is a point on the circumference of the circle, the prove that ∠PMQ = 90° by vector method.

Given: X is the origin. PMQ is a semi-circle. PQ is the diameter of a semi-circle. To prove: ∠PMQ = 90°
Construction: Join X to M.
Proof:
Statement Reason
i) $\overrightarrow{QM}$ = $\overrightarrow{QX} + \overrightarrow{XM}$ = $\overrightarrow{XM} - \overrightarrow{XQ}$ i) Triangle law of vector addition.
ii) $\overrightarrow{MP}$ = $\overrightarrow{MX} + \overrightarrow{XP}$
= $\overrightarrow{XP} - \overrightarrow{XM}$
ii) Triangle law of vector addition.
iii) $\overrightarrow{QM}$ = $\overrightarrow{XM} - (-\overrightarrow{XP})$ = $\overrightarrow{XM} + \overrightarrow{XP}$ iii) $\overrightarrow{XP} = -\overrightarrow{XQ}$
iv) $\overrightarrow{QM}$.$\overrightarrow{MP}$ = $(\overrightarrow{XM} + \overrightarrow{XP}).(\overrightarrow{XP} - \overrightarrow{XM})$ = $(\overrightarrow{XP})^2 - (\overrightarrow{XM})^2$ iv) Multiplying statement (ii) and (iii)
v) $\overrightarrow{QM}$.$\overrightarrow{MP}$ = $(\overrightarrow{XM})^2 - (\overrightarrow{XM})^2$ = 0 v) XP = XM (radius)
vi) ∴ ∠PMQ = 90° vi) Dot product of $\overrightarrow{PM}$ and $\overrightarrow{MQ}$ is zero.
Hence, It is proved that ∠PMQ is a right-angle.

In the given figure, PQ = PR and QS = RS, then prove by vector method: PS ⊥ QR.
OR, Prove by vector method that the line joining the mid-point of the base and the vertex of an isosceles triangle is perpendicular to the base.

Given: In △PQR, PQ = PR and QS = RS To prove: PS ⊥ QR
Proof:

Statement Reason
i) $\overrightarrow{PS}$ = $\frac{1}{2}(\overrightarrow{PQ} + \overrightarrow{PR})$ i) Mid-point theorem of vector.
ii) $\overrightarrow{QR}$ = $\overrightarrow{PR}$ - $\overrightarrow{PQ}$ ii) Triangle law of vector addition.
iii) $\overrightarrow{PS}$.$\overrightarrow{QR}$ = $\frac{1}{2}(\overrightarrow{PQ} + \overrightarrow{PR})(\overrightarrow{PR}$ - $\overrightarrow{PQ})$ iii) Multiplying statement (i) and (ii).
iv) $\overrightarrow{PS}$.$\overrightarrow{QR}$ = $\frac{1}{2}(\overrightarrow{PQ})^{2} - (\overrightarrow{PR})^2$ iv) From statement (iii)
v) $\overrightarrow{PS}$.$\overrightarrow{QR}$ = $\frac{1}{2}(\overrightarrow{PQ})^{2} - (\overrightarrow{PQ})^2$ v) PQ = PR (given)
vi) $\overrightarrow{PS}$.$\overrightarrow{QR}$ = 0 vi) From statement (v)
vii) PS ⊥ QR vii) Dot product of $\overrightarrow{PS}$ and $\overrightarrow{QR}$ is zero.
Hence, It is proved that PS ⊥ QR.

In the given figure, ∠RPQ = 90° and S is the mid-point of RQ. Prove by vector method that: PS = QS = RS.

Given: In right angled triangle PQR, ∠RPQ = 90° and S is the mid-point of RQ (QS = SR). To prove: PS = QS = RS
Proof:
Statement Reason
i) ∠RPQ = 90° i) △PQR is a right angled triangle.
ii) $\overrightarrow{PR}$.$\overrightarrow{PQ}$ = 0 ii) ∠RQP = 90°
iii) $(\overrightarrow{PS} + \overrightarrow{SR}).(\overrightarrow{PS}$ + $\overrightarrow{SQ})$ = 0 iii) Triangle law of vector addition.
iv) $(\overrightarrow{PS} + \overrightarrow{SR}).(\overrightarrow{PS}$ - $\overrightarrow{QS})$ = 0 iv) $\overrightarrow{SQ}$ = -$\overrightarrow{QS}$
v) $(\overrightarrow{PS} + \overrightarrow{SR}).(\overrightarrow{PS}$ - $\overrightarrow{SR})$ = 0
or, $(\overrightarrow{PS})^2$ - $(\overrightarrow{SR})^2$ = 0
or, $(\overrightarrow{PS})^2$ = $(\overrightarrow{SR})^2$
v) QS = SR (given)
vi) PS = SR vi) From (v)
vii) QS = RS vii) S is the mid-point of QR.
viii) PS = QS = RS viii) From statement (vi) and (vii)
Hence, It is proved that PS = QS = RS.

In the given figure, PQ = QR = RS = SP. Prove by vector method: PR ⊥ QS.

Let, $\overrightarrow{PQ}$ = $\overrightarrow{SR}$ = $\overrightarrow{b}$ and $\overrightarrow{QR}$ = $\overrightarrow{PS}$ = $\overrightarrow{a}$ Given: PQ = QR = RS = SP
To prove: PR ⊥ QS
Statement Reason
i) $\overrightarrow{PR}$.$\overrightarrow{QS}$ = $(\overrightarrow{PQ} + \overrightarrow{QR}).(\overrightarrow{QR} + \overrightarrow{RS})$ i) Triangle law of vector addition
ii) $\overrightarrow{PR}$.$\overrightarrow{QS}$ = $(\overrightarrow{PQ} + \overrightarrow{QR}).(\overrightarrow{QR} - \overrightarrow{SR})$ ii) $\overrightarrow{RS}$ = -$\overrightarrow{SR}$
iii) $\overrightarrow{PR}$.$\overrightarrow{QS}$ = $(\overrightarrow{a} + \overrightarrow{b}).(\overrightarrow{a} - \overrightarrow{b})$
= $(\overrightarrow{a})^2$ - $(\overrightarrow{b})^2$
iii) Replacing the value that we considered above
iv) $\overrightarrow{PR}$.$\overrightarrow{QS}$ = $(\overrightarrow{a})^2$ - $(\overrightarrow{a})^2$ = 0 iv) Given
v) PR ⊥ QS v) Dot product of $\overrightarrow{PR}$ and $\overrightarrow{QS}$ is zero.
Hence, it is proved that PR ⊥ QS.

If $\overrightarrow{OP}$ = $\overrightarrow{p}$, $\overrightarrow{OQ}$ = $\overrightarrow{q}$ and the point M divides the line segment PQ internally in the ratio of m1 : m2 then prove that: $\overrightarrow{OM}$ = $\frac{m_{1}\overrightarrow{q} + m_{2}\overrightarrow{p}}{m_1 + m_2}$

Given: $\overrightarrow{OP}$ = $\overrightarrow{p}$, $\overrightarrow{OQ}$ = $\overrightarrow{q}$ and the point M divides the line segment PQ internally in the ratio of m1 : m2.

Let, $\overrightarrow{OM}$ = $\overrightarrow{m}$

To prove: $\overrightarrow{OM}$ = $\frac{m_{1}\overrightarrow{q} + m_{2}\overrightarrow{p}}{m_1 + m_2}$
Proof:

Statement Reason
i) $\overrightarrow{PM}$ = $\overrightarrow{PO}$ + $\overrightarrow{OM}$ = $\overrightarrow{m}$ - $\overrightarrow{p}$ i) Triangle law of vector addition.
ii) $\overrightarrow{MQ}$ = $\overrightarrow{MO}$ + $\overrightarrow{OQ}$ = $\overrightarrow{q}$ - $\overrightarrow{m}$ ii) Same as above.
iii) $\frac{\overrightarrow{PM}}{\overrightarrow{MQ}}$ = $\frac{m_1}{m_2}$ iii) Given
iv) $\frac{\overrightarrow{m} - \overrightarrow{p}}{\overrightarrow{q} - \overrightarrow{m}}$ = $\frac{m_1}{m_2}$ iv) From statement (i), (ii), and (iii).
v) m2$\overrightarrow{m}$ - m2$\overrightarrow{p}$ = m1$\overrightarrow{q}$ - m1$\overrightarrow{m}$
or, m2$\overrightarrow{m}$ + m1$\overrightarrow{m}$ = m2$\overrightarrow{p}$ + m1$\overrightarrow{q}$
so, $\overrightarrow{m}$ = $\frac{m_{1}\overrightarrow{q} + m_{2}\overrightarrow{p}}{m_1 + m_2}$
v) From statement (iv)
Hence, it is proved that $\overrightarrow{OM}$ = $\frac{m_{1}\overrightarrow{q} + m_{2}\overrightarrow{p}}{m_1 + m_2}$.

If $\overrightarrow{OA}$ = $\overrightarrow{a}$, $\overrightarrow{OB}$ = $\overrightarrow{b}$ and the point M divides the line segment AB externally in the ration of m : n then prove that: $\overrightarrow{OM}$ = $\frac{m\overrightarrow{b} - n\overrightarrow{a}}{m + n}$

Given: $\overrightarrow{OA}$ = $\overrightarrow{a}$, $\overrightarrow{OB}$ = $\overrightarrow{b}$ and the point M divides the line segment AB externally in the ration of m : n

To prove: $\overrightarrow{OM}$ = $\frac{m\overrightarrow{b} - n\overrightarrow{a}}{m + n}$

Proof:

Statement Reason
i) $\overrightarrow{AM}$ = $\overrightarrow{AO}$ + $\overrightarrow{OM}$ = $\overrightarrow{m}$ - $\overrightarrow{a}$ i) Triangle law of vector addition.
ii) $\overrightarrow{BM}$ = $\overrightarrow{BO}$ + $\overrightarrow{OM}$ = $\overrightarrow{m}$ - $\overrightarrow{b}$ ii) Same as above.
iii) $\frac{\overrightarrow{AM}}{\overrightarrow{BM}}$ = $\frac{m}{m}$ iii) Given
iv) $\frac{\overrightarrow{m} - \overrightarrow{a}}{\overrightarrow{m} - \overrightarrow{b}}$ = $\frac{m}{n}$ iv) From statement (i), (ii), and (iii).
v) n$\overrightarrow{m}$ - n$\overrightarrow{a}$ = m$\overrightarrow{m}$ - m$\overrightarrow{b}$
or, m$\overrightarrow{m}$ - n$\overrightarrow{m}$ = m$\overrightarrow{b}$ - n$\overrightarrow{a}$
so, $\overrightarrow{m}$ = $\frac{m\overrightarrow{b} - n\overrightarrow{a}}{m - n}$
v) From statement (iv)
Hence, it is prove that $\overrightarrow{OM}$ = $\frac{m\overrightarrow{b} - n\overrightarrow{a}}{m - n}$.

Prove by vector method that the diagonals of a rectangle are equal.

Let PQRS is a rectangle. PR and QS are its diagonals. To prove: PR = QS
Proof:
Statement Reason
i) $\overrightarrow{PR}$ = $\overrightarrow{PQ}$ + $\overrightarrow{QR}$ = $\overrightarrow{QR}$ - $\overrightarrow{QP}$ i) Triangle law of vector addition.
ii) $\overrightarrow{PR}$ = $\overrightarrow{QR}$ - $\overrightarrow{RS}$ ii) Opposite side of rectangle is equal.
iii) $(\overrightarrow{PR})^2$ = $(\overrightarrow{QR}$ - $\overrightarrow{RS})^2$
= $(\overrightarrow{QR})^2$ - 2$\overrightarrow{QR}$.$\overrightarrow{RS}$ + $(\overrightarrow{RS})^2$
iii) Squaring statement (ii)
iv) $(\overrightarrow{PR})^2$ = $(\overrightarrow{QR})^2$ + $(\overrightarrow{RS})^2$ iv) ∠QRS = 90°, so, $\overrightarrow{QR}$.$\overrightarrow{RS}$ = 0.
v) $\overrightarrow{QS}$ = $\overrightarrow{QR}$ + $\overrightarrow{RS}$ v) Same as statement (i)
vi) $(\overrightarrow{QS})^2$ = $(\overrightarrow{QR})^2$ + 2$\overrightarrow{QR}$.$\overrightarrow{RS}$ + $(\overrightarrow{RS})^2$ vi) Squaring statement (v)
vii) $(\overrightarrow{QS})^2$ = $(\overrightarrow{QR})^2$ + $(\overrightarrow{RS})^2$ Same as statement (iv)
viii) $(\overrightarrow{PR})^2$ = $(\overrightarrow{QS})^2$
so, PR = QS
viii) From statement (iv) and statement (vii)
Hence, It is proved that PR = QS.

In the given triangle PQR, ∠QPR = 90° and S is a mid-point of QR, prove by the vector method that: SP = SQ = SR.

In the given triangle XYZ, ∠XYZ = 90° and A is the mid-point of XZ. Prove by vector method that: XA = YA = ZA.

Prove by vector method that the median of an isosceles triangle is perpendicular to the base.

In the given figure, PQRS is a rectangle. Prove by vector method that: PR = QS.

Prove by vector method that, the diagonals of rhombus intersect each other at right angle.

By using vector method, prove that the quadrilateral formed by joining the midpoints of adjacent sides of a quadrilateral is a parallelogram.

Given: ABCD is quadrilateral in which midpoints of sides AB, BC, CD, DA are P, Q, R, S respectively. To prove: PQRS is parallelogram.
Construction: Join A and C.
Proof:
Statement Reason
i) $\overrightarrow{PQ}$ = $\frac{1}{2} \overrightarrow{AC}$ and $\overrightarrow{PQ} \parallel \overrightarrow{AC}$ i) In △ABC, the lines segment joining the midpoint of two sides of triangle is half of the third side and parallel to third side.
ii) $\overrightarrow{SR}$ = $\frac{1}{2} \overrightarrow{AC}$ and $\overrightarrow{SR} \parallel \overrightarrow{AC}$ ii) In ▵ADC, the line segment joining the midpoint of two sides of triangle is half of third side and parallel to third side.
iii) $\overrightarrow{PQ}$ = $\overrightarrow{SR}$ and $\overrightarrow{PQ} \parallel \overrightarrow{SR}$ iii) From statement (i) and (ii).
iv) PQRS is parallelogram iv) From statement (iii), being opposite side of quadrilateral equal and parallel.
Hence, It is proved that PQRS is parallelogram.

Prove by vector method that the diagonals of a parallelogram bisect each other.

Let O be the origin, OABC is a parallelogram and M and N are midpoints of AC and OB respectively. To prove: $\overrightarrow{OM}$ = $\overrightarrow{ON}$
Proof:
Statement Reason
i) $\overrightarrow{OM}$ = $\frac{1}{2} (\overrightarrow{OA} \, + \, \overrightarrow{OC})$ i) Midpoint theorem.
ii) $\overrightarrow{ON}$ = $\frac{\overrightarrow{OB}}{2}$ ii) N is the midpoint of AB.
iii) $\overrightarrow{ON}$ = $\frac{\overrightarrow{OA} \, + \, \overrightarrow{AB}}{2}$ iii) Triangle law of vector addition.
iv) $\overrightarrow{ON}$ = $\frac{\overrightarrow{OA} \, + \, \overrightarrow{OC}}{2}$ iv) $\overrightarrow{AB}$ = $\overrightarrow{OC}$, in parallelogram.
v) $\overrightarrow{OM}$ = $\overrightarrow{ON}$ v) From statement (i) and (iv).
Hence, It is proved that diagonals of parallelogram bisect to each other.