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Since L_x=12, L_y=3 and L_z=4 are given, we can find out L by

L ₂= L_x² + L_y² + L_z²

=12² + 3² + 4²

=144 + 9 + 16 = 169

Hence L=13 units.

1. (d) (1, -1, 1)

2. (a) \(\cfrac2{\sqrt3}\) units

3. (c) \(\cfrac{x-1}1=\cfrac{y}{-1}=\cfrac{z-1}1\)

4. (d) Both (a) and (c)

5. (b) \(\Big(\cfrac{1}{\sqrt3},\cfrac{-1}{\sqrt3},\cfrac{1}{\sqrt3}\Big)\)

1. (b) 3x + 2y + 4z = 11

2. (a) \(\cfrac{5}{\sqrt{29}}\) units

3. (c) \(\cfrac{x-2}3=\cfrac{y-3}2=\cfrac{z-1}4\)

4. (d)\(\Big(\cfrac{43}{29},\cfrac{77}{29},\cfrac{9}{29}\Big)\)

5. (b) ) \(\cfrac{\sqrt{29}}2\) sq. units

We have

vector AB = (1 + 2)i + (2 - 3)j + (3 - 5)k = 3i - j - 2k

vector BC = (7 - 1)i + (0 - 2)j + (-1 - 3)k = 6i - 2j - 4k

vector CA = (7 + 2)i + (0 - 3)j + (-1 - 5)k = 9i - 3j - 6k

Now, |vector AB|² = 14, |vector BC|² = 56, |vector CA|² = 126

⇒ |vector AB| = √14, |vector BC| = 2√14, |vector CA| = 3√14

⇒ |vector CA| = | vector AB| + |vector BC|

Hence the points A, B and C are collinear.

(i) |\(\bar a\)| = |3i + j + 2k| = \(\sqrt{14}\)

(ii) Since projection of \(\bar a\) on another vector \(\bar b\) and magnitude of \(\bar a\) is \(\sqrt{14}\), then \(\bar a\) and \(\bar b\) are parallel,

(b) 6i + 2j + 4k.

(iii) Projection of \(\bar a\) on \(\bar c\)

= |\(\bar a\)|cos60° = \(\sqrt{14} \times \frac{1}{2}\) = \(\frac{\sqrt{14}}{2}.\)

\(\bar{a}\) = 3i + 4j, \(\bar{b}\) = 4i - 3j

\(|\bar{a}|\) = \(\sqrt{9 + 16}\) = 5,

\(|\bar{b}|\) = \(\sqrt{16 + 9}\) = 5

\(\bar {a} \;and \; \bar {b}\) are different but \(|\bar{a}|\) =\(|\bar{b}|\)

1. 15, 0, 0 : 0, 8, 6

2. 15i+0j+0k,  2: 0i+8j+6k

3. 15 unit , \(\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}\) = 10 unit

4.  \(\vec N=\vec A \times\vec B\) 

\(N=\begin{vmatrix}i&j&k\\15&0&0\\0&8&6\end{vmatrix}\) =15(6j - 8k) = -90j +120k ;  -90 , 120

5. \(\sqrt{(-90)^2+120^2}=\sqrt{8100+14400}=\sqrt{22500}\) = 150

Answer of second part: \(\vec F\)= 910 (1/2 \(\hat i\)-6/7 \(\hat j\) +1/7 \(\hat k\)) = 455 \(\hat i\)– 780 \(\hat j\) + 130 \(\hat k\).

The dot product is just \(\vec F.\vec N\)= 455*(0) -780*(-90) + 130*120 = 85,800 watts.

From the definition of dot product: \(\vec F.\vec N\) = |\(\vec F\)||\(\vec N\)|cosθ

Then since | \(\vec F\)| = 910 and |\(\vec N\)| = 150 and \(\vec F.\vec N\)= 85,800 we have

cosθ = (85800/(910 x150)) = 0.629 and so θ =cos^-1 (0.629) which is 0.8905 rad and is 51° .(using cosine table)

This is the angle between the normal to the surface and the incident solar rays.

The compliment of this is the elevation of the sun above the plane of the roof or 90 - 51 = 39°. .

1. (a) |\(\vec a+\vec b\) |² = |\(\vec a-\vec b\)|² ⇒ 2.\(\vec a.\vec b\)= 0, \(\vec a\perp\vec b\)

2. (a) 0

3. (b) 2 \(sin\cfrac{\theta}2\)

4. (c) ±2(\(\vec b\times\vec c\)) 

5. (c) √70/2 sq units

\(\vec{a} + \vec{b} + \vec{c}\)= \(\hat{i}\)(1 - 2 + 1) + \(\hat{j}\)(-2 + 4 - 6) + \(\hat{k}\)(1 + 5 - 7)

= - \(4\hat{j} - \vec{k}\)

\((\hat{i}\times \hat{j}). \hat{k}+(\hat{j}\times \hat{k}).\hat{i}\) \(=\hat{k}.\hat{k}+\hat{i}.\hat{i}\)

\(=1+1=2\)

[Note: \(\vec{a}.\vec{b}=|\vec{a}|.|\vec{b}|\,cos\,\theta.\) Also \(|\hat{i}|=|\hat{j}|=|\hat{k}|=1\)]

Required area of parallelogram \(=|2\hat{i}\times 3\hat{j}|\)

\(=6|\hat{i}\times \hat{j}|=6|\hat{k}|=6\) sq. units.

\(\big[\)Note : Area of parallelogram whose sides are represented by \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a}\times \vec{b}|\)\(\big]\)

Let A be the point (1, 0, 0) and B be the point (0, 1, 0) (i.e.,) vector OA = i and vector OB = j 

Then vector AB = vector (OB - OA) = j - i = - i + j 

= (-1, 1, 0) 

= (a, a + b, a + b + c) 

⇒ a = -1, a + b = 1 and a + b + c = 0 

Now a = -1 ⇒ -1 + b = 1 ;a + b + c = 0

⇒ b = 2; -1 + 2 + c = 0 ⇒ c + 1 = 0 

⇒ c = -1 

∴ a = -1; b = 2; c = -1. 

Note: If we taken vector BA then we get a = 1, b = -2 and c = 1.

Let vector β₁ = λ vector α, λ is a scalar, i.e. vector β₁ = 3λi - λj.

Now, vector β₂ = vector(β - β₁) = (2 - 3λ)i + (1 + λ)j - 3k

Now, since vector β₂ is perpendicular vector α, we should have vector (α x β) = 0 i.e.,

3(2 - 3λ) - (1 + λ) = 0

⇒ 6 - 9λ - 1 - λ = 0 ⇒ 5 - 10λ = 0

⇒ 10λ = 5 ⇒ λ = 5/10 = 1/2

Therefore vector β₁ = 3/2i - 1/2j and vector β₂ = 1/2i + 3/2j - 3k

i. Collinear

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors. Two collinear vectors may point in either same or opposite direction. But, they cannot be inclined at some angle from each other.

Hence FE (\(\vec{\text x}\)), AD (\(\vec{\text z}\)) and BC (\(\vec{\text b}\)) are collinear vectors.

And also AF (\(\vec{\text y}\)) and CD ( \(\vec{\text c}\)) are collinear vectors.

And AB (\(\vec{\text a}\) ) and ED (\(\vec{\text d}\)) are collinear vectors.

ii. Equal

Equal vectors are vectors that have the same magnitude and the same direction. Equal vectors may start at different positions.

Hence AF (\(\vec{\text y}\)) and CD (\(\vec{\text c}\) ) are equal vectors.

And also FE (\(\vec{\text x}\)) and BC (\(\vec{\text b}\)) are equal vectors.

And AB (\(\vec{\text a}\)) and ED (\(\vec{\text d}\)) are equal vectors.

iii. Co - initial

Any given two vectors are called co - initial vectors if both the given vectors have the same initial point.

Hence, AB (\(\vec{\text a}\)), AF (\(\vec{\text y}\)) and AD (\(\vec{\text z}\)) are co - initial vectors.

iv. Collinear but not equal.

And AD ( \(\vec{\text z}\)) and BC ( \(\vec{\text b}\)) are collinear but not equal vectors.

And AD (\(\vec{\text z}\)) and FE (\(\vec{\text x}\)) are collinear but not equal vectors.

i. Time period - is a scalar quantity as it involves only magnitude. A scalar quantity is a one - dimensional measurement of a quantity. Eg: 10 seconds has only magnitude, i.e., 10 and no direction.

ii. Distance - is a scalar quantity as it involves only magnitude. A scalar quantity is a one dimensional measurement of a quantity. Eg: 5meters has only magnitude 5 and no direction. 

iii. Displacement - is vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction.

iv. Force - is a vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction. Eg., 5N downward has magnitude of 5 and direction is downward.

v. Work done - is a scalar quantity as it involves only magnitude and no particular direction. A scalar quantity is a one dimensional measurement of a quantity.

vi. Velocity - is a vector quantity as it involves both magnitude as well as direction. Vector quantity has both magnitude and direction. Eg., 5m/s east has magnitude of 5m/s and also direction towards east.

vii. Acceleration is a vector quantity because it involves both magnitude as well as direction.

(i) 10 kg is a scalar quantity because it involves only magnitude.
(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.
(iii) 40° is a scalar quantity as it involves only magnitude.
(iv)40 watts is a scalar quantity as it involves only magnitude.
(v) 10^–19 coulomb is a scalar quantity as it involves only magnitude.
(vi)20 m/s² is a vector quantity as it involves magnitude as well as direction.

Scalar quantity: Work 

Vector quantity: Force, velocity, displacement.

(i) Coinitial \(\bar{b} \;and \;\bar{a}\) are co-initial vectors 

(ii) Equal \(\bar{b} \;and \;\bar{d}\) are equal vectors 

(iii) Collinear but not equal \(\bar{a} \;and \;\bar{c}\) are collinear but not equals

(i) Time period is a scalar quantity as it involves only magnitude.
(ii) Distance is a scalar quantity as it involves only magnitude.
(iii) Force is a vector quantity as it involves both magnitude and direction.
(iv) Velocity is a vector quantity as it involves both magnitude as well as direction.
(v) Work done is a scalar quantity as it involves only magnitude.

(i) time period – Scalar 

(ii) distance – Scalar 

(iii) force – Vector 

(iv) velocity – Vector 

(v) work done – Scalar

(i) 10 kg – Scalar 

(ii) 2 meters north-west – Vector 

(iii) 40° – Scalar 

(iv) 40 watt – Scalar 

(v) 10^-19 coulomb- Scalar 

(vi) 20 m/s² – Vector

i. 15 kg - is a scalar quantity as this involves only mass. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

ii. 20 kg weight - is a vector quantity as it involves both magnitude and direction. Weight is a force which is a vector and has a magnitude and direction.

iii. 45° is a scalar quantity as it involves the only magnitude. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

iv. 10 meters south - east is a vector quantity as it involves both magnitude and direction. v. 50 m/sec² is a scalar quantity as it involves a magnitude of acceleration. A scalar quantity is a one - dimensional measurement of a quantity.

sin θ = cos θ

or   θ = 45°

Given  \(\vec a\)is a vector and m is a scalar such that m\(\vec a\) = \(\vec 0\)

Let \(\vec a\) = \(a_1\hat i+b_1\hat j + c_1\hat k\) 

then according to the given question

m\(\vec a\) = \(\vec 0\)

⇒ \(m(a_1\hat i+b_1\hat j+c_1\hat k)\) = \(0\hat i+0\hat j+0\hat k\)

⇒ \((ma_1\hat i+mb_1\hat j+mc_1\hat k)\) = \(0\hat i+0\hat j+0\hat k\)

Compare the coefficients of \(\hat i, \hat j,\hat k\), we get

ma₁ = 0 ⇒ m = 0 or a₁ = 0

Similarly, mb₁ = 0 ⇒ m = 0 or b₁ = 0

And, mc₁ = 0 ⇒ m = 0 or c₁ = 0

From the above three conditions,

m = 0 or a₁ = b₁ = c₁ = 0

⇒ m = 0 or \(\vec a\) = \(a_1\hat i+b_1\hat j + c_1\hat k\) = \(0\hat i+0\hat j+0\hat k\) = 0

Hence the alternatives for m and \(\vec a\) are m = 0 or  \(\vec a\) = 0

Given: \(|\vec a|=|\vec b|\)

It means the magnitude of the vector \(\vec a\) is equal to the magnitude of the vector \(\vec b\), but we cannot conclude anything about the direction of the vector.

And we know that \(\vec a = \vec b\) means magnitude and same direction. So, it is false that\(|\vec a|=|\vec b|\) ⇒ \(\vec a=\vec b\)

Given: \(|\vec a| = |\vec b|\)

It means the magnitude of the vector \(\vec a\) is equal to the magnitude of the vector \(\vec b\) , but we cannot conclude anything about the direction of the vector.

So it is false that \(|\vec a|=|\vec b|\) ⇒ \(\vec a= \pm\vec b\)

Given vectors a = 2i + 3j - k and b = i - 2j + k.

Let vector c  be the resultant vector a  and vector b  then

vector c = (2i + 3j - k) + (i - 2j + k) = 3i + j + 0k

⇒ | vector c| = √(9 + 1 + 0) = √10

∴ Unit vector in the direction of vector c = c = 1/|vector c|(vector c) = 1/√10(3i + j)

Hence, the vector of magnitude 5 units and parallel to the resultant of vectors a and vector b  is

± 5c = ± 5(1/√10)(3i + j) = ±(3√10/2)i ±(√10/2)j 

We know that two nonzero vectors are perpendicular if their scalar product is zero.

Here, vector(a + b) = 5i - j - 3k + i + 3j - 5k = 6i + 2j - 8k

and vector (a - b) = 5i - j - 3k - i - 3j + 5k = 4i - 4j + 2k

Now, vector(a + b) x vector (a - b) = (6i + 2j - 8k) x (4i - 4j + 2k)  = 24 - 8 - 16 = 0

Hence vector(a + b) and vector (a - b) are perpendicular.

The angle θ between two vectors a  and vector b  is given by

cosθ = (vector a x vector b)/(|vector a||vector b|)

Now, vector(a.b) = (i + j - k) x (i - j + k) = 1 - 1 - 1 = - 1

Therefore, we have cosθ = -1/3 ⇒ θ = cos^-1(-1/3)

Position vectors of points A, B and C are respectively given as

vector a = 3i - 4j - 4k, vector b = 2i - j + k and vector c = i - 3j - 5k

Now, vector (AB) = vector b - vector a = 2i - j + k - 3i + 4j = - i + 3j + 5k

⇒ |vector AB|² = 1 + 9 + 25 = 35

vector BC = vector c - vector b = i - 3j - 5k - 2i + j - k = - i - 2j - 6k

⇒ |vector BC|² = 1 + 4 + 36 = 41

vector CA = vector a - vector c = 3i - 4j - 4k - i + 3j + 5k = 2i - j + k

⇒ |vector CA|² = 4 + 1 + 1 = 6

⇒ |vector BC|² = |vector AB|² + |vector CA|²

Hence it form the vertices of a right angled triangle.

i x (j + k) + j x (k + i) + k(i + j)

= i x j + i x k + j x k + j x  i + k x i + k x j

= k - j + i - k + j - i = 0

Vector PQ =  - ()i + (5 - 3)j + (6 - 0)k 

= 3i + 2j + 6k

∴ Required unit vector 

= (3i + 2j + 6k)/√(9 + 4 + 36) 

= (3i + 2j + 6k)/√49 = (3i + 2j + 6k)/7 

= 3/7i + 2/7j + 6/7k

Required vector 

= 21((2i - 3j + 6k)/√(4 + 9 + 36)) = 21((2i - 3j + 6k)/√49)

= 21((2i - 3j + 6k)/7) = 3(2i - 3j + 6k) = 6i - 9j + 18k

1. \(\overline {AB}\) = 2i + j + 2k

2. \(|\overline {AB}| = \sqrt{4+1+4}\) =3

The direction cosines are \(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\).

3. cos α = \(\frac{2}{3}\) ⇒ α = cos^-1(\(\frac{2}{3}\)).

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vector (a+b+c)=(1-2+1)i+(-2+4-6)j+(1+5-7)k

= -4j+k

\(\bar a\) + \(\bar b\) = 3i + 4j + k + 2i – 7j – 3k = 5i – 3j – 2k
\(\bar a\) − \(\bar b\) = 3i + 4j + k – (2i – 7j -3k) = i + 11j + 4k
\(\bar b\) + \(\bar c\) = 2i – 7j -3k + 2i +3j – 9k
= 4i – 4j – 12k.

Given that vectors a and vector b be such that |vector a| = 3 and |vector b| = √2/3.

Also, vector a x vector b is  a unit vector ⇒ |vector a x vector b| = 1

⇒ |vector a| x |vector b|sin = 1 ⇒ 3 x √2/3sinθ = 1

⇒ sinθ = 1/2 ⇒ θ = π/4

Since in an equilateral triangle, orthocenter and centroid coincide, therefore the position vector of centroid is  \(\vec 0.\)

Also, the position vector of centroid G( \(\vec g\)) can be defined as \(\cfrac{\vec a+\vec b+\vec c}3\)

Therefore, \(\cfrac{\vec a+\vec b+\vec c}3\) = \(\vec 0\)  hence  \(\vec a+\vec b+\vec c=\vec 0\)

(d) coplanar vectors

(b) 60° 

α = 60°, β = 45° 

We know cos² α + cos² β + cos² γ = 1

(i.e.,) (1/2)² + (1/√2)² + cos² γ = 1

cos² γ = 1 - 1/4 - 1/2 = 1/4

cos γ = 1/2 ⇒ y = π/3 = 60°

(c) 2 vector(2a + b)/3 

The sum of squares of direction cosines of a vector is 1.

Let the angles made by vector be α, β, γ.

Then, we get cos²α+cos²β+cos²γ=1

using cos²θ=1 - sin²θ, we get

(1 - sin²α) +(1 - sin²β) + (1 - sin²γ) = 1

Or, sin²α + sin²β + sin²γ = 2