Samacheer Class 12 Maths - Applications of Vector Algebra

Let O be origin, A,B endpoints of chord with position vectors a,b. Midpoint M has m=(a+b)/2. Chord vector AB = b-a. OM·AB = m·(b-a)=((a+b)/2)·(b-a) = (1/2)(a·b - a·a + b·b - b·a) = (1/2)(|b|^2-|a|^2). But A and B lie on same circle so |a|=|b|, hence OM·AB=0. Therefore OM ⟂ AB.

Let triangle with base endpoints A,B (position vectors a,b) and apex C (c) with |c-a|=|c-b|. Midpoint M of AB has m=(a+b)/2. CM·AB = (c-m)·(b-a) = c·(b-a) - m·(b-a). But m·(b-a)=((a+b)/2)·(b-a)=(1/2)(|b|^2-|a|^2)=0 if we choose origin conveniently or note symmetry. Alternatively compute (c-a)·(c-a)=(c-b)·(c-b) ⇒ c·c -2c·a + a·a = c·c -2c·b + b·b ⇒ 2c·(b-a)=b·b-a·a. Using midpoint gives (c-m)·(b-a)= (1/2)(2c·(b-a) - (b·b-a·a))=0. Hence CM ⟂ AB.

Take rhombus with adjacent side vectors a and b from origin; vertices O, a, a+b, b. Diagonals are d1 = a+b (from O to opposite vertex) and d2 = b-a (from a to b). Their midpoints coincide at (a+b)/2 so they bisect each other. Their dot product: (a+b)·(b-a)=|b|^2-|a|^2=0 since |a|=|b| in a rhombus. Hence diagonals are perpendicular.

Let diagonals intersect at O. Write AC = u and BD = v. Quadrilateral splits into 4 triangles; two opposite triangles have area (1/2)| (u/2)×(v/2)| etc. More directly, area = area(ΔAOB)+ΔBOC+ΔCOD+ΔDOA = (1/2)(| (u/2)×(v/2)| + | (u/2)×(v/2)| + | (u/2)×(v/2)| + | (u/2)×(v/2)| ) = 4*(1/2)*(1/4)|u×v| = (1/2)|u×v|. Thus area = (1/2)|AC×BD|.

Let base vector be b and the other side vectors be a and a' which are parallel (so a'=a+λb). Area of parallelogram with sides b and a is |b×a| and with b and a' is |b×a'| = |b×(a+λb)| = |b×a + λ(b×b)| = |b×a| since b×b=0. Hence areas equal.

$\overline { F_1 }$ = -3$\hat { i }$ + 6$\hat { j }$ – 3$\hat { k }$
$\overline { F_2 }$ = 4$\hat { i }$ – 10$\hat { j }$ + 12$\hat { k }$
$\overline { F_3 }$ = 4$\hat { i }$ + 7$\hat { j }$
$\overline { F }$ = $\overline { F_1 }$ + $\overline { F_2 }$ + $\overline { F_3 }$
= 5$\hat { i }$ + 3$\hat { j }$ + 9$\hat { k }$
$\overline { OB }$ = 8$\hat { i }$ – 6$\hat { j }$ – 4$\hat { k }$
$\overline { OA }$ = 18$\hat { i }$ + 3$\hat { j }$ – 9$\hat { k }$
$\overline { AB }$ = $\overline { OB }$ – $\overline { OA }$
= -10$\hat { i }$ – 9$\hat { j }$ + 5$\hat { k }$
$\overline { t }$ = $\overline { r }$ × $\overline { F }$ = $\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-10 & -9 & 5 \\
5 & 3 & 9
\end{array}\right|$
= $\hat { i }$(-81 – 15) – $\hat { j }$(-90 – 25) + $\hat { k }$(-30 + 45)
= -96$\hat { i }$ + 115$\hat { j }$ + 15$\hat { k }$

$\overline { a }$.($\overline { b}$ × $\overline { c }$) = [ $\overline { a }$, $\overline { b }$, $\overline { c }$ ] = $\left|\begin{array}{ccc}
1 & -2 & 3 \\
2 & 1 & -2 \\
3 & 2 & 1
\end{array}\right|$
= 1(1 + 4) + 2(2 + 6) + 3(4 – 3)
= 5 + 16 + 3 = 24

Volume of the parallelepiped = [ $\overline { a }$, $\overline { b }$, $\overline { c }$ ]
= $\left|\begin{array}{ccc}
-6 & 14 & 10 \\
14 & -10 & -6 \\
2 & 4 & -2
\end{array}\right|$
= -6(20 + 24) -14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760
= 720 cu. units.

If $\overline { a }$, $\overline { b }$ and $\overline { c }$ are coplanar [ $\overline { a }$, $\overline { b }$ and $\overline { c }$ ] = 0
$\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 0 & 0 \\
c_{1} & c_{2} & c_{3}
\end{array}\right|$ = 0
1(0) – 1(c 3 ) + 1(c 2 ) = 0
-c 3 + c 2 = 0
c 3 = c 2 = 2

[ $\overline { a }$, $\overline { b }$ and $\overline { c }$ ] = $\left|\begin{array}{ccc}
1 & 0 & -1 \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|$
= 1(1 + x – y – x + x²)-1(x² – y)
(1 + x – y – x + x² – x² + y)
= 1
There is no x and y terms
∴ [ $\overline { a }$, $\overline { b }$ and $\overline { c }$ ] depends on neither x nor y.

[ $\overline { a }$, $\overline { b }$ and $\overline { c }$ ] = 0
$\left|\begin{array}{lll}
a & a & c \\
1 & 0 & 1 \\
c & c & b
\end{array}\right|$ = 0
a(0 – c) – a(b – c) + c(c) = 0
-ac – ab + ac + c² = 0
c² – ab = 0
c² = ab
⇒ c in the geometric mean of a and b.

Hence proved

Hence proved.

(i) Points are (2, 1, 4) (a – 1, 4, -1)
Cartesian equation

(ii) Points are (0, 2, b – 1) (5, 3, -2)
Cartesian equation

$\overline { a }$ = -2$\hat { i }$ + 3$\hat { j }$ + 4$\hat { k }$ (x 1, y 1, z 1 ) = (-2, 3, 4)
$\overline { b }$ = -4$\hat { i }$ + 5$\hat { j }$ + 6$\hat { k }$ (l, m, n) = (-4, 5, 6)
vector equation

If the lines are perpendicular
b 1 d 1 + b 2 d 2 + b 3 d 3 = 0
(5m + 2) – 5(2m) + 1(+3) = 0
5m + 2 – 10m + 3 = 0
-5m = -5
m = $\frac { -5 }{ -5 }$ = 1
⇒ ∴ m = 1

Given points (5, 2, 8)

2$\hat { i }$ + $\hat { j }$ + 3$\hat { k }$ is a vector perpendicular to both the given straight lines it passes through (5, 2, 8)
The equation is $\overline { r }$ = (5$\hat { i }$ + 2$\hat { j }$ + 8$\hat { k }$) + t(2$\hat { i }$ + $\hat { j }$ – 2$\hat { k }$)
Cartesian form = $\frac { x-5 }{ 2 }$ = $\frac { y-2 }{ 1 }$ = $\frac { z-8 }{ -2 }$

$\overline { b }$ is not a scalar multiple of $\overline { d }$
∴ They are not parallel.
∴ The given lines are skew lines.

$\overline { r }$ = $\overline { a }$ + t$\overline { b }$
$\overline { a }$ = -i + 3j + k, $\overline { b }$ = 2i + 3j – k
Given points D (5, 4, 2) to the point A. If P is the foot of the perpendicular from to the straight line.
F is of the form
(2t – 1, 3t + 3, -t + 1) and
$\overline { DF }$ = $\overline { OF }$ – $\overline { OD }$ = (2t – 6)i + (3t – 1)j + (-t – l)k
$\overline { b }$ is perpendicular to $\overline { DF }$, we have
$\overline { b }$.$\overline { DF }$ = 0 => (2t – 6) 2 + 3(3t – 1) – 1(-t – 1) = 0
4t- 12 + 9t – 3 + t + 1 = 0
14t – 14 = 0
14t = 14
t = 1
∴ F (2 – 1, 3 + 3, -1 + 1) = F (1, 6, 0) is foot point. Equation of the perpendicular.
(x 1, y 1, z 1 ) = (5, 4, 2), (x 2, y 2, z 2 ) = (1, 6, 0).

Normal vector n = (3,4,−5) has magnitude |n| = √(9+16+25)=√50=5√2. A plane at distance d from origin with normal n has equation r·(n/|n|)=±d ⇒ r·n = ±d|n|. Here d=7 so r·(3,4,−5)=±7·5√2 = ±35√2. Thus vector equations: r·(3i+4j−5k)=35√2 or r·(3i+4j−5k)=−35√2. Cartesian forms: 3x+4y−5z = ±35√2.

If normal makes equal acute angles with axes, its direction cosines are (l,l,l) with l>0 and l^2+l^2+l^2=1 ⇒ 3l^2=1 ⇒ l=1/√3. Unit normal = (1/√3,1/√3,1/√3). Given magnitude = 3√3, the normal vector is n = (3√3)(1/√3,1/√3,1/√3) = 3(1,1,1). Plane through P(−1,1,2): n·(r−r0)=0 ⇒ 3(1,1,1)·(x+1,y−1,z−2)=0 ⇒ x+1 + y−1 + z−2 =0 ⇒ x+y+z = 2.

Let intercepts be a,b,c. Centroid = (a/3, b/3, c/3) = (u,v,w) ⇒ a=3u, b=3v, c=3w. Therefore plane intercept form: x/a + y/b + z/c = 1 ⇒ x/(3u) + y/(3v) + z/(3w) =1. Multiply by 3: x/u + y/v + z/w = 3.

Vector along given two points: v = (9−2,3−2,6−1) = (7,1,5). Given plane normal m = (2,6,6). The plane sought is perpendicular to the plane with normal m, so its normal n must be perpendicular to m and also perpendicular to no — actually the plane must contain the line through the two given points and be perpendicular to the given plane; hence its normal is n = v × m. Compute n = (7,1,5) × (2,6,6) = (−24,−32,40) = −8(3,4,−5). Using point (2,2,1): (3,4,−5)·(x−2,y−2,z−1)=0 ⇒ 3x+4y−5z = 9. Vector form: r·(3,4,−5) = 9.

Cartesian equation
-12x + 11y + 16z = 14
12x – 11y – 16z = -14
12x – 11y – 16z + 14 = 0

Non parametric form

Which is the required Cartesian equation of the place.

Cartesian equation
9x – 2y – 5z = -4
9x – 2y – 5z + 4 = 0

Take A(3,6,2). Direction vectors: AB = B−A = (−4,−4,4), AC = C−A = (3,−2,−4). Parametric: r = A + s AB + t AC. Normal n = AB × AC = (24,−4,20) = 4(6,−1,5) so take n = (6,−1,5). Cartesian: n·(r − A) = 0 ⇒ 6(x−3) −1(y−6) +5(z−2)=0 ⇒ 6x − y +5z −22 = 0. Non‑parametric vector form: (r − A)·(6,−1,5)=0.

Cartesian equation:
3x + Sy – 7z = 6
3x + 5y – 7z – 6 = 0

Take $P=(2,3,4),\ \vec u=(1,1,3)$ and $Q=(1,4,5),\ \vec v=(-3,2,1)$. Then $\vec{PQ}=(-1,1,1)$ and $\vec u\times\vec v=(-5,-10,5)$. Since $\vec{PQ}\cdot(\vec u\times\vec v)=5-10+5=0$, the lines are coplanar. The plane has normal $(1,2,-1)$ (i.e. $\vec u\times\vec v$ divided by $-5$) and passes through $(2,3,4)$: $1(x-2)+2(y-3)-1(z-4)=0$, i.e. $x+2y-z=4$.

2(4 – m 4 ) – 2(m² – 2) = 0
8 – 2m 4 – 2m² + 4 = 0
12 – 2m 4 – 2m² = 0
(÷ -2) -6 + m 4 + m² = 0
m 4 + m² – 6 = 0
(m² – 2)(m² + 3) = 0
m² – 2 = 2; m² = -3 (not possible)
m² = 2
m = ±√2

If the two lines are coplanar

When λ = 2
(x 1, y 1, z 1 ) = (1, -1, 0)
(b 1, b 2, b 3 ) = (2, 2, 2)
(d 1, d 2, d 3 ) = (5, 2, 2)
$\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
b_{1} & b_{2} & b_{3} \\
d_{1} & d_{2} & d_{3}
\end{array}\right|$ = 0
⇒ $\left|\begin{array}{ccc}
x-1 & y+1 & z-0 \\
2 & 2 & 2 \\
5 & 2 & 2
\end{array}\right|$ = 0
⇒ (x – 1)(0) – (y + 1)(-6) + z(6) = 0
⇒6(y + 1) – 6z = 0
⇒ 6y + 6 – 6z = 0
⇒ y – z + 1 = 0
When λ = 2
(b 1, b 2, b 3 ) = (2, -2, 2)
(d 1, d 2, d 3 ) = (5, 2, -2)
⇒ $\left|\begin{array}{ccc}
x-1 & y+1 & z-0 \\
2 & -2 & 2 \\
5 & 2 & -2
\end{array}\right|$ = 0
⇒ (x – 1)(0) – (y + 1)(-14) + z(4 + 10) = 0
⇒ 14(y + 1) + 14z = 0
⇒ 14y + 14 + 14z = 0
⇒ y + z + 1 = 0

Given planes are
$\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})$ = 3
2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0
Equation of a plane which passes through the line of intersection of the planes
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0 …………… (1)
This passes through the point (-2, 1, 3).
(1) ⇒ (-4 – 7 + 12 – 3) + λ(-6 – 5 + 12 + 11) = 0
-2 + λ(12) = 0 ⇒ 12λ = 2
λ = $\frac{2}{12}$ ⇒ λ = $\frac{1}{6}$
The required equation is
(1) ⇒ (2x – 7y + 4z – 3) + $\frac{1}{6}$ (3x – 5y + 4z + 11) = 0
12x – 42y + 24z – 18 + 3x – 5y + 4z + 11 = 0
15x – 47y + 28z – 7 = 0

Required equation of the plane
(x + 2y + 3z – 2) + λ(x – y + z – 3) = 0 ………. (1)
(1 + λ)x + (2 – λ)y + (3 + λ)z + (-2 – 3λ) = 0 ………(2)
Distance from (2) to the point (3, 1, -1) is $\frac { 2 }{ √3 }$

putting
λ = $\frac { -7 }{ 2 }$ in (1)
The required equation
(x + 2y + 3z – 2) – $\frac { 7 }{ 2 }$ (x – y + z – 3) = 0
2x + 4y + 6z – 4 – 7x + 7y – 7z + 21 = 0
-5x + 11y – z + 17 = 0
5x – 11y + z – 17 = 0

For a plane parallel to 2x−3y+5z−7=0 the normal is (2,−3,5). Equation through (3,4,1): 2(x−3) −3(y−4) +5(z−1)=0 ⇒ 2x−3y+5z +1 = 0. Distance between planes |(−7) − (1)|/√(4+9+25) = 8/√38.

Using the distance formula, length $=\dfrac{|1-2+3-5|}{\sqrt{1^2+(-1)^2+1^2}}=\dfrac{|-3|}{\sqrt3}=\dfrac{3}{\sqrt3}=\sqrt3$ units.

Any point on the line x – 1 = $\frac{y}{2}$ = z + 1 is
x – 1 = $\frac{y}{2}$ = z + 1 = λ,(say)
(λ + 1, 2λ, λ – 1)
This passes through the plane 2x – y + 2z = 2
2(λ + 1) – 2λ + 2(λ – 1) = 2
2λ + 2 – 2λ + 2λ – 2 = 2
λ = 1
∴ The required point of intersection is (2, 2, 0)

The normal to the plane is $\vec n=(1,2,3)$, with $|\vec n|^2=1+4+9=14$. For $P=(4,3,2)$, $\vec n\cdot P=4+6+6=16$, so $\vec n\cdot P-2=14$. The foot of the perpendicular is $Q=P-\dfrac{\vec n\cdot P-2}{|\vec n|^2}\,\vec n=(4,3,2)-\dfrac{14}{14}(1,2,3)=(3,1,-1)$. The length of the perpendicular is $PQ=\dfrac{|\vec n\cdot P-2|}{|\vec n|}=\dfrac{14}{\sqrt{14}}=\sqrt{14}$ units.

(c) –$\frac { 1 }{ 2 }$, -2
Hint: