Vector Algebra Class 12 Important Questions with Solutions Previous Year Questions
Algebra of Vectors
Question 1.
Find the position vector of a point which divides the join of points with position vectors
Answer:
Let given position vectors are
Let
∴
= 4
Question 2.
If
Answer:
Given vectors are
Now,
= 6î – 3ĵ + 2k̂
and
=
∴ The unit vector parallelto the vector
Question 3.
The two vectors ĵ + k̂ and 3î – ĵ + 4k̂ represent the two sides
Answer:
Given,
Alternate Method:
Given
Question 4.
Write the direction ratios of the vector 3
Answer:
Clearly, 3
= (3î + 3ĵ – 6k̂) + (4î – 8ĵ + 10k̂)
= 7î – 5ĵ + 4k̂
Hence, direction ratios of vectors 3
Question 5.
Find the unit vector in the direction of the sum of the vectors 2î + 3ĵ – k̂ and 4î – 3ĵ + 2k̂. (Foreign 2015)
Answer:
Let
Now, sum of two vectors,
Question 6.
Find a vector in the direction of vector 2î – 3ĵ + 6k̂ which has magnitude 21 units. (Foreign 2014)
Answer:
To find a vector in the direction of given vector, first of all we find unit vector in the direction of given vector and then multiply it with given magnitude.
Let
Then, |
=
The unit vector in the direction of the given vector
Now, the vector of magnitude equal to 21 units
and in the direction of a is given by
21â = 21
Question 7.
Find a vector a of magnitude 5√2, making an angle of
Answer:
Here, we have l = cos
⇒ l =
Question 8.
Write a unit vector in the direction of the sum of the vectors
Answer:
Question 9.
Find the value of p for which the vectors 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are parallel. (All India 2014)
Answer:
Given, 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are two parallel vectors, so their direction ratios will be proportional.
Question 10.
Write the value of cosine of the angle which the vector
Answer:
Given,
Now, unit vector in the direction of
∴ Cosine of the angle which given vector makes with Z-axis is
Question 11.
Find the angle between X-axis and the vector î + ĵ + k̂. (All India 2014C)
Answer:
Let
Now, unit vector in the direction of
So, angle between X-axis and the vector
î + ĵ + k̂ is cos α =
[∵ â = lî + mĵ + nk̂ and cos α = l ⇒ α = cos-1l]
Question 12.
Write a vector in the direction of the vector î – 2ĵ + 2k̂ that has magnitude 9 units. (Delhi 2014C)
Answer:
3î – 6ĵ + 6k̂
Question 13.
Write a unit vector in the direction of vector
Answer:
First, find the vector
Given points are P (1, 3, 0) and Q (4, 5, 6).
Here, x1 = 1, y1 = 3, z1 = 0
and x2 = 4, y2 = 5, z2 = 6
So, vector PQ = (x2 – x1)k̂ + (y2 – y1)ĵ + (z2 – z1)k̂
= (4 – 1)î + (5 – 3)ĵ + (6 – 0)k̂
= 3î + 2ĵ + 6k̂
∴ Magnitude of given vector
Hence, the unit vector in the direction of
Question 14.
If a unit vector
Answer:
Here, we have
l = cos
Question 15.
Write a unit vector in the direction of the sum of vectors
Answer:
Question 16.
If
Answer:
Two vectors are equal, if coefficients of their components are equal.
Given,
On comparing the coefficient of components, we get
x = 3, y = -2, z = -1
Now, x + y + z = 3 – 2 – 1 = 0
Question 17.
P and Q are two points with position vectors 3
Answer:
Question 18.
L and M are two points with position vectors 2
Answer:
5
Question 19.
A and B are two points with position vectors 2
Answer:
Given, A and B are two points with position vectors 2
Question 20.
Find the sum of the vectors
Answer:
Given vectors are
Sum of the vectors
= – 4ĵ – k̂
Question 21.
Find the sum of the following vectors.
Answer:
3î – ĵ – 2k̂
Question 22.
Find the sum of the following vectors.
Answer:
5î – 5ĵ + 3k̂
Question 23.
Find the scalar components of
Answer:
Given initial point is A (2,1) and terminal point is B (- 5, 7), then scalar component of
x2 – x1 = – 5 – 2 = – 7and y2 – y1 = 7 – 1 = 6.
Question 24.
For what values of
Answer:
If
Let given vectors are
We know that, vectors
∴ 2î – 3ĵ + 4k̂ = k(aî + 6ĵ – 8k̂)
On comparing the coefficients of î and ĵ, we get
2 = ka and -3 = 6k ⇒ k = –
∴ 2 = –
Question 25.
Write the direction cosines of vector -2î + ĵ – 5k̂. (Delhi 2011)
Answer:
Direction cosines of the vector aî + bĵ + ck̂ are
Question 26.
Write the position vector of mid-point of the vector joining points P(2, 3, 4) and Q (4, 1, – 2). (Foreign 2011)
Answer:
Mid-point of the position vectors
Given points are P(2, 3, 4) and Q(4,1,-2) whose position vectors are
Now, position vector of mid-point of vector joining points P(2, 3, 4) and Q(4, 1, – 2) is
Question 27.
Write a unit vector in the direction of vector
Answer:
We know that, unit vector in the direction of â is â =
Required unit vector in the direction of vector
Question 28.
Find the magnitude of the vector
Answer:
Magnitude of a vector r = xî + yĵ + zk̂ is |
Given vector is a = 3i – 2/ + 6fc.
∴ Magnitude of
=
Question 29.
Find a unit vector in the direction of vector
Answer:
Question 30.
If A, B and C are the vertices of a ΔABC, then what is the value of
Answer:
Let ΔABC be the given triangle.
Now, by triangle law of vector addition,
Question 31.
Find a unit vector in the direction of
Answer:
Question 32.
Find a vector in the direction of
Answer:
4î – 2ĵ + 4k̂
Question 33.
Find the position vector of mid-point of the line segment AB, where A is point (3, 4, -2) and Bis point (1, 2, 4). (Delhi 2010)
Answer:
2î + 3ĵ + k̂
Question 34.
Write a vector of magnitude 9 units in the direction of vector -2î + ĵ + 2k̂. (All India 2010)
Answer:
-6î + 3ĵ + 6k̂
Question 35.
Write a vector of magnitude 15 units in the direction of vector î – 2ĵ + 2k̂. (Delhi 2010)
Answer:
5î – 10ĵ + 10k̂
Question 36.
What is the cosine of angle which the vector √2î + ĵ + k̂ makes with Y-axis? (Delhi 2010)
Answer:
Question 37.
Find a vector of magnitude 5 units and parallel to the resultant of
Answer:
First, find resultant of the vectors a and o, which is
Given,
Now, resultant of above vectors =
= (2î + 3ĵ – k̂) + (î – 2ĵ + k̂) = 3î + ĵ
Question 38.
Let
Answer:
First, find the vector 2 a – b + 3c, then find a unit vector in the direction of 2a-b + 3c.
After this, the unit vector is multiplying by 6.
Given,
∴
= 2 (î + ĵ + k̂) – (4î – 2ĵ + 3k̂) + 3 (î – 2ĵ + k̂)
= 2î + 2ĵ + 2k̂ – 4î + 2ĵ – 3k̂ + 3î – 6ĵ + 3k̂
⇒
Now, a unit vector in the direction of vector
Hence, vector of magnitude 6 units parallel to the Vector
= 2î – 4ĵ + 4k̂
Question 39.
Find the position vector of a point R, which divides the line joining two points P and Q whose position vectors are 2
Answer:
Given,
and
Let OR be the position vector of point R, which divides PQ in the ratio 1 : 2 externally
Now, we have to show that P is the mid-point of RQ,
Hence, P is the mid-point of line segment RQ.
Product of Two Vectors and Scalar Triple Product
Question 1.
Find the magnitude of each of the two vectors
Answer:
Question 2.
Find the value of [î, k̂, ĵ], (CBSE 2018C)
Answer:
[î, k̂, ĵ] = î ∙ (k̂× ĵ)
= -[[î, k̂, ĵ] = –
Question 3.
Find λ and μ, if (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂) = 0. (All India 2016)
Answer:
Given, (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂)
= î(3μ + 9λ) ĵ k̂
On comparing the coefficients of î, ĵ and k̂ , we get
3μ + 9λ = 0, – μ + 27 = 0 and – λ – 9 = 0
⇒ μ = 27 and – λ = 9
⇒ μ = 27 and λ = – 9
Also, the values of μ and λ satisfy the equation
3μ + 9λ = 0.
Hence, μ = 27 and λ = – 9.
Question 4.
Write the number of vectors of unit length perpendicular to both the vectors
Answer:
We know that, unit vectors perpendicular to
and
So, there arc two unit vectors perpendicular to the given vectors.
Question 5.
If
Answer:
Question 6.
If
Answer:
Question 7.
Find λ, if the vectors
Answer:
⇒ 1(- 3 + λ) – 3(6) + 1(2λ) = 0
[expanding along R1]
⇒ – 3 + λ – 18 + 2λ = 0
⇒ 3λ = 21
∴ λ = 7
Question 8.
If
Answer:
Question 9.
If â, b̂ and ĉ are mutually perpendicular unit vectors, then find the value of |2â + b̂ + ĉ |. (All India 2015)
Answer:
Given â, b̂ and ĉ are mutually perpendicular unit vectors, i.e.
Question 10.
Write a unit vector perpendicular to both the vectors
Answer:
First, determine perpendicular vectors of
Given vector are
As we know the, vectors
Then,
= î(0 -1) – ĵ(0 – 1) + k̂(1 – 1)
= – î + ĵ
Then , the unit vector perpendicular to both
Question 11.
Find the area of a parallelogram whose adjacent sides are represented by the vectors 2 î – 3 k̂ and 4 ĵ + 2 k̂. (Foreign 2015)
Answer:
Let adjacent sides of a parallelogram bc
Question 12.
If
Answer:
Question 13.
If
Answer:
Question 14.
Find the projection of the vector î + 3ĵ + 7k̂ on the vector 2î – 3 ĵ + 6k̂. (Delhi 2014)
Answer:
let
Question 15.
Write the projection of vector î + ĵ + k̂ along the vector ĵ. (Foreign 2014)
Answer:
1
Question 16.
Write the value of the following. î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ). (Foreign 2014)
Answer:
we have, î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ)
= î × ĵ + î × k̂ × ĵ × k̂ + ĵ × î + k̂ × î + k̂ × ĵ
[∵ cross product is distributive over addition]
= k̂ – ĵ + î – k̂ + ĵ – î =
[∵ î × ĵ = k̂, î × k̂ = – ĵ, ĵ × k̂ = î, ĵ × î = – k̂, k̂ × î = ĵ, k̂ × ĵ = – î ]
Question 17.
If vectors
Answer:
Question 18.
Find
Answer:
Given,
= 2(4 – 1) – 1 (- 2 – 3) + 3( – 1 – 6)
= 2 × 3 – 1 × (-5) + 3 × (- 7)
= 6 + 5 – 21 = 11 – 21 = – 10
Question 19.
If
Answer:
Question 20.
If |
Answer:
let θ be the angle between latex]\vec{a}[/latex] and
Question 21.
Write the projection of the vector
Answer:
Question 22.
Write the value of λ, so that the vectors a = 2î + λĵ + k̂ and b = î – 2ĵ + 3k̂ are perpendicular to each other. (Delhi 2013C, 2008)
Answer:
Given vectors are
and
Since, vectors are perpendicular.
∴
⇒ (2î + λĵ + k̂) ∙ (î – 2ĵ + 3k̂)
⇒ 2 – 2λ + 3 = 0
∴ λ = 5/2
Question 23.
Write the projection of (
Answer:
Question 24.
Write the projection of the vector 7î + ĵ – 4k̂ on the vector 2î + 6 ĵ + 3k̂. (Delhi 2013C)
Answer:
Question 25.
If
Answer:
Question 26.
Find |
Answer:
Given,
Question 27.
Find λ, when projection of
Answer:
Given,
⇒ 2λ + 18 = 28
⇒ 2λ = 10
∴ λ = 5
Question 28.
Write the value of (k̂ × ĵ) . î + ĵ . k̂. (All India 2012)
Answer:
Use the results k̂ × ĵ = – î
ĵ ∙ k̂ and î ∙ î = 1 and simplify it.
Given, (k̂ × î) ∙ î + ĵ ∙ k̂ = (- î) ∙ î + ĵ ∙ k̂
= – (î ∙ î) + 0 = – 1 [∵ (î ∙ î) = 1]
Question 29.
If
Answer:
From Eqs. (i) and (ii). it may be concluded that
Question 30.
Write the projection of vector î – ĵ on the vector î + ĵ. (All India 2011)
Answer:
0
Question 31.
Write the angle between vectors
Answer:
let θ be the angle between
cos θ =
Question 32.
For what value of λ are the vectors î + 2λĵ + k̂ and 2î + ĵ – 3k̂ perpendicular? (All India 2011C)
Answer:
Question 33.
If |
Answer:
Question 34.
Find the value of λ, if the vectors 2î + λĵ + 3k and 3î + 2ĵ – 4k̂ are perpendicular to each other. (All India 2010c)
Answer:
3
Question 35.
If |
Answer:
Question 36.
If
Answer:
Use the following formulae:
and
where, θ is the angle between
Question 37.
Find λ, if (2î + 6ĵ + 14k̂) × (î – λĵ + Ik̂) = 0. (All India 2010)
Answer:
– 3
Question 38.
If the sum of two unit vectors a and b is a unit vector, show that the magnitude of their difference is √3. (Delhi 2019, 2012c)
Answer:
let
[taking positive square root, as magnitude cannot be negative]
Question 39.
If
Answer:
= 2(- 4 – 1) – 3(2 + 3) + 1(1 – 6)
= – 10 – 15 – 5 = – 30
Question 40.
If |
Answer:
let θ be the angle between
Question 41.
Find the volume of cuboid whose edges are given by -5î + 7ĵ + 5k̂, -5î + 7ĵ – 5k̂ and 7î – 5 ĵ – 5k̂. (All India 2019)
Answer:
= |- 3 (- 21 – 15) – 7 (15 + 21) + 5(25 – 49)|
= |1108 – 252 – 120|
= 264 cubic units
Question 42.
Show that the points A(-2î + 5ĵ + 5k̂), B(î + 2 ĵ + 5k̂) and C(7î – k̂) are collinear. (All India 2019)
Answer:
Question 43.
Find
Answer:
We have,
∴
= î ( – 2 – 15) – ĵ (- 4 – 9) + k̂(10 – 3)
= – 17î + 13ĵ + 7k̂
Question 44.
If θ is the angle between two vectors î – 2 ĵ + 3k̂ and 3î – 2 ĵ + k̂, find sin θ. (CBSE 2018)
Answer:
let
Question 45.
If
Answer:
(5)2 + 2 × 5 × 6 × cos θ + (6)2 = (9)2
⇒ 25 + 60 cos θ + 36 = 81
⇒ cos θ =
⇒ θ = cos-1
Question 46.
If î + ĵ + k̂, 2î + 5ĵ, 5î + 2ĵ – 5k̂ and î – 6ĵ – k̂ respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether
Answer:
Question 47.
The scalar product of the vector
Answer:
Question 48.
Let
Answer:
We have,
Since,
= λ[î(5 – 4) – ĵ(15 + 1) + k̂(- 12 – 1)]
= λ(î – 16ĵ – 13k̂)
Also, it is given that
∴ λ(î – 16ĵ – 13k̂) ∙ (4î + 5ĵ – k̂) = 21
⇒ λ(4 – 80 + 13) = 21
⇒ λ(- 63) = 21
⇒ λ =
Now from Eq. (j), we get
Question 49.
Find x such that the four points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D (7, 7, 2) are coplanar. (CBSE 2018C)
Answer:
Given points are A(4, 4, 4), B (5, x, 8), C(5, 4, 1) and D(7, 7, 2), then position vectors of A, B, C and D respectively, are
⇒ 1(0 + 9) – (x – 4) (- 2 + 9) + 4(3 – 0) = 0
⇒ 9 – (x – 4) (7) + 12 = 0
⇒ 9 – 7x + 28 + 12 = 0
⇒ 49 – 7x = 0
⇒ 7x = 49
⇒ x = 7
Question 50.
Find the value of x such that the points A(3, 2, 1), B(4, x, 5), C(4, 2,- 2) and D (6, 5, -1) are coplanar. (All India 2017)
Answer:
5
Question 51.
If
Answer:
If three vectors
cos θ =
Question 52.
Using vectors, find the area of the ΔABC, whose vertices are A(1, 2, 5), 5(2, -1, 4) and C(4, 5, -1). (Delhi 2017; All India 2013)
Answer:
Let the position vectors of the verices A, B and C of ΔABC be
Question 53.
Let
(a) Let c1 = 1 and c2 = 2, find c3 which makes
(b) If c2 = – 1 and c3 = 1, show that no value of c1 can make
Answer:
Given,
The given vectors are coplanar iff
(a) If c1 = 1 and c2 = 2,
Then, from Eq.(i), we get
⇒ – 1(c3 – 0) + 1(2 – 0) = 0
⇒ – c3 + 2 = 0
⇒ – c3 = – 2
⇒ c3 = 2
(b) If c2 = – 1 and c3 = 1, then from Eq. (i), we get
⇒ 1(0) – 1(1 – 0) + 1(- 1 – 0) = 0
⇒ 0 – 1 – 1 = 0
⇒ – 2 ≠ 0
∴ No value of c1 can make 1’ and coplanar.
Hence proved
Question 54.
Show that the points A, B, C with position vectors 2î – ĵ + k̂, î – 5ĵ – 5k̂ and 5î – 4ĵ – 4k̂ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle. (All India 2017)
Answer:
Question 55.
Show that the vectors
Or
Prove that, for any three vectors
Answer:
Question 56.
Show that the four points A (4, 5, 1), B(0, -1, -1), C(3, 9, 4) and D (-4, 4, 4) are coplanar. (All India 2016)
Or
Show that the four points A, B, C and D with position vectors 4î + 5ĵ + k̂, – ĵ – k̂, 3î + 9ĵ + 4k̂ and 4(- î + ĵ + k̂), respectively are coplanar. (All India 2014)
Answer:
Let the position vector of points A, B, C and D are
= – 4(12 + 3) + 6 (- 3 + 24) – 2(1 + 32)
= – 60 + 126 – 66 = 0
Hence, the four points A, B, C and D are coplanar.
Question 57.
The two adjacent sides of a parallelogram are 2î – 4ĵ – 5k̂ and 2î + 2 ĵ + 3k̂. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram. (All India 2016)
Answer:
Let ABCD be the given parallelogram with
Question 58.
If
Answer:
Use the result, if two vectors are parallel, then their cross-product will be a zero vector.
Question 59.
If
Answer:
Question 60.
If
Answer:
Question 61.
Find the value of λ so that the four points A, B,C and D with position vectors 4 î + 5ĵ + k̂, -ĵ – k̂,3i + Xj+4k and – 4 î + 4ĵ + 4 k̂, respectively are coplanar. (Delhi 2015C)
Answer:
Use the condition that four points with position vectors
On expanding along R1, we get
⇒ – 4(3λ – 15 + 3) + 6(- 3 + 24) – 2(1 + 8λ – 40) = 0
⇒ – 4(3λ – 12) + 6(21) – 2(8λ – 39) = 0
⇒ – 12λ + 48 + 126 – 16λ + 78 = 0
⇒ – 28λ + 252 = 0
λ = 9
Question 62.
Prove that
Answer:
Question 63.
If
Answer:
Question 64.
Vectors
Answer:
Question 65.
The scalar product of the vector
Or
The scalar product of vector i + j + k with the unit vector along the sum of vectors 2î + 4ĵ – 5k̂ and λî + 2ĵ + 3k̂ is equal to one. Find the value of λ. (All India 2009,2008C)
Answer:
First, determine the unit vector of
⇒ (λ + 6)2 = λ2 + 4λ + 44 [squaring both sides]
⇒ λ2 + 36 + 12λ + 4λ + 44
⇒ 8λ = 8
⇒ λ = 1
Hence, the value of λ is 1.
On substituting the value of λ in Eq. (1), we get Unit vector along
Question 66.
Find the vector
Answer:
= î (25 – 4) – ĵ (20 + 1) + k̂(- 16 – 5)
= î(21) – ĵ(21) + k̂(- 21)
= 21î – 21ĵ – 21k̂
So,
Also, given that
∴ (21Î»î – 21λĵ – 21λk̂) . (3î + ĵ – k̂) = 21
⇒ 63λ – 21λ + 21λ = 21
⇒ 63λ = 21
⇒ λ =
On putting λ =
which is the required vector.
Question 67.
Find the unit vector perpendicular to both of the vectors
Answer:
⇒ 2x + 3y + 4z = 0 …… (ii)
and (xî + yĵ + zk̂) . (- ĵ – 2k̂) = 0
⇒ – y – 2z = 0
⇒ y = – 2z
On putting the value of yin Eq. (ii), we get
2x + 3 (- 2z) + 4z = 0
⇒ x = z
On substituting the value of x and y in Eq. (1),
we get
⇒ z2 + 4z2 + z2 = 1
⇒ 6z2 = 1
⇒ z = ±
then, x = ±
and y = ±
Hence, the required vectors are
Question 68.
Find the unit vector perpendicular to the plane ABC where the position vectors of A, B and C are 2î – ĵ + k̂, î + ĵ + 2k̂ and 2î + 3k̂, respectively. (All India 2014C)
Answer:
A unit vector perpendicular to plane ABC is
Let O be the origin of reference.
Question 69.
Dot product of a vector with vectors î – ĵ + k̂, 2î + ĵ – 3k̂ and î + ĵ + k̂ are respectively 4, 0 and 2. Find the vector. (Delhi 2013C)
Answer:
⇒ a1 + a2 + a3 = 2
On subtracting Eq. (iii) from Eq. (i), we get
– 2a2 = 2
⇒ a2 = – 1
On substituting a2 = – 1 in Eq. (ii) and (iii),
we get
2a2 – 3a3 = 1 …… (iv)
⇒ a1 + a3 = 3
On multiplying Eq. (v) by 3 and then adding with Eq. (iv), we get
5a1 = 1 + 9 = 10
⇒ a1 = 2
On substituting a1 = 2 in Eq. (v), we get
a3 = 1
Hence, the vector is
Question 70.
Find the values of λ for which the angle between the vectors
Answer:
let θ be the obtuse angle between the vectors
14λ2 – 7λ < 0
= 2λ2 – λ < 0
Either λ < 0, 2λ – 1 > 0 or λ > 0, 2λ – 1 < 0
= Either λ < 0, λ >
0 < λ <
λ ∈
Question 71.
If a, b and c are three vectors such that each one is perpendicular to the vector obtained by sum of the other two and |
Or
If
Answer:
Question 72.
If
Answer:
Given
Let
Consider,
⇒ (x1 + x2)î + (y1 + y2)ĵ + (z1 + z2)k̂ = 2î + ĵ – 3k̂
On comparing the coefficient of î ĵ and k̂ both sides, we get
x1 + x2 = 2
y1 + y2 = 1
z1 + z2 = -3
Now, consider
⇒
⇒ x1 = 3λ, y1 = -λ,and z1 = 0 …(iv)
On substituting the values of x, y and z, from Eq. (iv) to Eq. (i), (ii) and (iii), respectively, we get
x2 = 2- 3λ, y2 = -1 + λ and z2 = -3 …(v)
Since, b2 ± a , therefore b2 a = 0
⇒ 3x2 – y2 = 0
⇒ 3 (2 – 3λ) – (1 + λ) = 0
⇒ 6 – 9λ – 1 – λ = 0
⇒ 5 – 10λ = 0
⇒ λ =
On substituting λ =
Question 73.
If
Answer:
Given
Let
= î (z – y) – ĵ(z – x) + k̂(y – x)
Now,
= î(z – y) + ĵ(x – z) + k̂(y – x)
= 0î + 1ĵ + (-1)k̂ [∵
On comparing the coefficients from both sides, we get
z – y = 0,x – z = 1, y – x = -1
⇒ y = z and x – y = 1…(i)
Also given,
⇒ (î + ĵ + k̂) . (xî + yĵ + zk̂) = 3
⇒ x + y + z = 3 (1)
⇒ x + 2y = 3 [∵ y = z] …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
3y = 2
⇒ y =
From Eq. (i),
x = 1 + y + 1 = 1 +
Hence,
Question 74.
If
Answer:
Use the result that if
Given,
Then,
= 6î – 2ĵ + (7 + λ) k̂
and a – = (î – ĵ + 7k̂) – (5î – j ̂+ λk̂)
= -4î + (7 – λ)k̂
Since,
vectors, then
⇒ [6î – 2ĵ + (7 + λ)k̂]- [-4î + (7 – λ)k̂] = 0 (1)
⇒ -24 + (7+ X)(7 – X) =0
⇒ 49 – λ2 = 24
⇒ λ2 = 25
∴ λ = ± 5
Question 75.
If p = 5î + λĵ – 3k̂ and q = î + 3ĵ – 5k̂, then find the value of λ, so that
Answer:
λ = ± 1
Question 76.
If
Answer:
-169
Question 77.
Let
Answer:
Given vectors are
and
Let
We have,
⇒ (xî + yĵ + zk̂) – (î + 4ĵ + 7k̂) = 0
⇒ x + 4y + 2z = 0 ………….(i)
and
⇒ (xî + yĵ + zk̂) . (3î – 2ĵ + 7k̂) = 0
⇒ 3x – 2y + 7z = 0 …(ii)
Also, given ~p-~c =18 (1)
⇒ (xî + yĵ + zk̂) . (2î – ĵ + 4k̂) = 0
⇒ 2x – y + 4z = 18 …(iii)
On multiplying Eq. (i) by 3 and subtracting it from Eq. (ii), we get
– 14y + z = 0 ..(iv)
Now, multiplying Eq. (i) by 2 and subtracting it from Eq. (iii), we get
– 9y = 18
⇒ y = -2
On putting y = -2 in Eq. (iv), we get
-14 (-2) + z = 0
⇒ 28 + z = 0
⇒ z = -28
On putting y = -2 and z = -28 in Eq. (i), we get
x + 4 (-2) + 2 (-28) = 0
⇒ x – 8 – 56 = 0
⇒ x = 64
Hence, the required vector is
i.e.
Question 78.
Find a unit vector perpendicular to each of the vectors
Answer:
Question 79.
If a and 6 are two vectors, such that |
Answer:
Question 80.
If vectors
Answer:
Given,
and
Also,
∴ (
Now,
⇒
Then, from Bq. (i), we get
[î (2 – λ) + ĵ (2 + 2λ) + k̂(3 + λ)].[3î + ĵ] = 0
⇒ 3(2 – λ) + 1(2+ 2k) = 0
⇒ 8 – λ = 0
∴ λ = 8
Question 81.
Using vectors, find the area of triangle with vertices A (1, 1, 2), 5(2, 3, 5) and C (1, 5, 5). (All India 2011)
Answer:
Question 82.
Using vectors, find the area of triangle with vertices A (2, 3, 5), B (3, 5, 8) and C(2, 7, 8). (Delhi 2010C)
Answer:
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