Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 10 Vector Algebra Miscellaneous Exercise Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 10 Vector Algebra Miscellaneous Exercise

Question 1.

Write down a unit vector in XY plane making an angle of 30° with the positive direction of x-axis.

Solution:

Let OX, OY, OZ be the coordinate axes. The line OP lies in the plane XY so that ∠XOP = 30°, ∠POY = 60° and ∠POZ = 90°.

∴ Direction angles of OP are 30°, 60° and 90°.

∴ Direction cosines of OP are cos 30°, cos 60° and cos 90°.

Question 2.

Find the scalar components and magnitude of the vector joining the points P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}).

Solution:

Positive vectors of P and Q are x_{1}\(\hat {i} \) + y_{1}\(\hat {j} \) + z_{1}\(\hat {k} \)

and x_{2}\(\hat {i} \) + y_{2}\(\hat {j} \) + z_{2}\(\hat {k} \) respectively.

∴ \(\overrightarrow{\mathrm{PQ}}\) = \(\overrightarrow{\mathrm{OQ}}\) – \(\overrightarrow{\mathrm{OP}}\)

= (x_{2}\(\hat {i} \) + y_{2}\(\hat {j} \) + z_{2}\(\hat {k} \)) – (x_{1}\(\hat {i} \) + y_{1}\(\hat {j} \) + z_{1}\(\hat {k} \))

= (x_{2} – x_{1})\(\hat {i} \) + (y_{2} – y_{1})\(\hat {j} \) + (z_{2} – z_{1})\(\hat {k} \)

∴ Scalar components of vector \(\overrightarrow{\mathrm{PQ}}\) are x_{2} – x_{1}, y_{2} – y_{1} and z_{2} – z_{1}

Further, |\(\overrightarrow{\mathrm{PQ}}\)| = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\).

Question 3.

A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops.

Determine the girl’s displacement from her initial point of departure.

Solution:

The girl walks 4 km westward. This is represented by vector \(\overrightarrow{\mathrm{OQ}}\) = 4\(\hat {i} \) …………….. (1)

Further she goes in the direction 30° east of north, i.e., she moves along PQ and stops at Q. PQ makes x-4 an angle of 60° with OP.

Scalar components of PQ along OX = PQ cos 60° = 3 cos 60° = \(\frac{3}{2}\).

Scalar vertical component of PQ along OY = 3 sin 60° = \(\frac{3 \sqrt{3}}{2}\).

∴ \(\overrightarrow{\mathrm{PQ}}\) = \(\frac{3}{2}\)\(\hat {i} \) + \(\frac{3 \sqrt{3}}{2}\)\(\hat {j} \) ……………… (2)

Thus, the girl walks along OP and then along PQ.

∴ \(\overrightarrow{\mathrm{OP}}\) + \(\overrightarrow{\mathrm{PQ}}\) =\(\overrightarrow{\mathrm{OQ}}\)

From (1) and (2),

Question 4.

If \(\vec {a}\) = \(\vec {b}\) + \(\vec {c}\), then is it true that |\(\vec {a}\)| = |\(\vec {b}\)| + |\(\vec {c}\)|?

Justify your answer.

Solution:

\(\vec {a}\) = \(\vec {b}\) + \(\vec {c}\)

|\(\vec {a}\)| = |\(\vec {b}\) + \(\vec {c}\)|

∴ |\(\vec {a}\)|^{2} = |\(\vec {b}\) + \(\vec {c}\)|^{2} = (\(\vec {b}\) + \(\vec {c}\)) . (\(\vec {b}\) + \(\vec {c}\))

= \(\vec {b}\).\(\vec {b}\) + \(\vec {b}\).\(\vec {c}\) + \(\vec {c}\).\(\vec {b}\) + \(\vec {c}\).\(\vec {c}\)

= |\(\vec {b}\)|^{2} + 2\(\vec {b}\).\(\vec {c}\) + |\(\vec {c}\)|^{2} [∵ \(\vec {b}\).\(\vec {c}\) = \(\vec {c}\).\(\vec {b}\)]

= |\(\vec {b}\)|^{2} + |\(\vec {c}\)|^{2} + 2|\(\vec {b}\)| |\(\vec {c}\)| cos θ,

where θ is the angle between \(\vec {b}\) and \(\vec {c}\).

When θ = 0° |\(\vec {a}\)|^{2} = |\(\vec {b}\)|^{2} + |\(\vec {c}\)|^{2} + 2|\(\vec {b}\)| |\(\vec {c}\)| = (|\(\vec {b}\)| + |\(\vec {c}\)|^{2}

⇒ |\(\vec {a}\)| = |\(\vec {b}\)| + |\(\vec {c}\)|

If θ ≠ 0°, |\(\vec {a}\)| ≠ |\(\vec {b}\)| + |\(\vec {c}\)|.

Question 5.

Find the value of x for which x (\(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \)) is a unit vector.

Solution:

Let \(\vec {a}\) = x(\(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \)) = x\(\hat {i} \) + x\(\hat {j} \) + x\(\hat {k} \)

|\(\vec {a}\)| = \(\sqrt{x^{2}+x^{2}+x^{2}}\) = \(\sqrt{3}\)x

But |x| = 1 ⇒ \(\sqrt{3}\)x = ± 1 ∴ x = ± \(\frac{1}{\sqrt{3}}\).

Question 6.

Find a vector of magnitude 5 units and parallel to the resultant of the vectors \(\vec {a}\) = 2\(\hat {i} \) + 3\(\hat {j} \) – \(\hat {k} \)

and \(\vec {b}\) = \(\hat {i} \) – 2\(\hat {j} \) + \(\hat {k} \).

Solution:

Let vector c be the resultant of \(\vec {a}\) and \(\vec {b}\).

Now, \(\vec {c}\) = \(\vec {a}\) + \(\vec {b}\)

= (2\(\hat {i} \) + 3\(\hat {j} \) – \(\hat {k} \)) + (\(\hat {i} \) – 2\(\hat {j} \) + \(\hat {k} \)) = 3\(\hat {i} \) + \(\hat {j} \)

⇒ |\(\vec {c}\)| = \(\sqrt{3^{2}+1^{2}\) = \(\sqrt{10}\).

Question 7.

If \(\vec {a}\) = \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \), \(\vec {b}\) = 2\(\hat {i} \) – \(\hat {j} \) + 3\(\hat {k} \)

and \(\vec {c}\) = \(\hat {i} \) – 2\(\hat {j} \) + \(\hat {k} \), find a unit vector parallel to the vector 2\(\vec {a}\) – \(\vec {b}\) + 3\(\vec {c}\).

Solution:

We have:

\(\vec {a}\) = \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \), \(\vec {b}\) = 2\(\hat {i} \) – \(\hat {j} \) + 3\(\hat {k} \)

\(\vec {c}\) = \(\hat {i} \) – 2\(\hat {j} \) + \(\hat {k} \)

∴ \(\vec {d}\) – 2\(\vec {a}\) – \(\vec {b}\) + 3\(\vec {c}\)

= 2(\(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \)) – (2\(\hat {i} \) – \(\hat {j} \) + 3\(\hat {k} \)) + 3(\(\hat {i} \) – 2\(\hat {j} \) + \(\hat {k} \))

= 2\(\hat {i} \) + 2\(\hat {j} \) + 2\(\hat {k} \) – 2\(\hat {i} \) + \(\hat {j} \) – 3\(\hat {k} \) + 3\(\hat {i} \) – 6\(\hat {j} \) + 3\(\hat {k} \)

= (2 – 2 + 3)\(\hat {i} \) + (2 + 1 – 6)\(\hat {j} \) + (2 – 3 + 3)\(\hat {k} \)

= 3\(\hat {i} \) – 3\(\hat {j} \) + 2\(\hat {k} \)

∴ |\(\vec {d}\)| = \(\sqrt{3^{2}+(-3)^{2}+2^{2}}\) = \(\sqrt{9+9+4}\) = \(\sqrt{22}\).

∴ Unit vector parallel to 2\(\vec {a}\) – \(\vec {b}\) + 3\(\vec {c}\) = \(\frac{3}{\sqrt{22}}\)\(\hat {i} \) – \(\frac{3}{\sqrt{22}}\)\(\hat {j} \) + \(\frac{2}{\sqrt{22}}\)\(\hat {k} \).

Question 8.

Show that the points A(1, – 2, – 8), B(5, 0, – 2) and C(11, 3, 7) are collinear and find the ratio in which B divides AC.

Solution:

The position vectors of A and C are \(\hat {i} \) – 2\(\hat {j} \) – 8\(\hat {k} \) and 11\(\hat {i} \) + 3\(\hat {j} \) + 7\(\hat {k} \) respectively.

Let the point B lying on AC divides it in ratio λ : 1

∴ Position vector of B is

⇒ A, B and C are collinear and B divides it in the ratio 2 : 3.

Alternate Method of collinearity:

So, A, B and C are collinear.

Question 9.

Find the position vector of a point R which divides the line segment joining the two points P and Q, whose position vectors are (2\(\vec {a}\) + \(\vec {b}\)) and (\(\vec {a}\) – 3\(\vec {b}\)),

externally in the ratio 1 : 2. Also, show that P is the middle point of line segment RQ.

Solution:

The position vector of P and Q are respectively (2\(\vec {a}\) + \(\vec {b}\)) and (\(\vec {a}\) – 3\(\vec {b}\)).

R divides PQ externally in the ratio 1 : 2.

⇒ P is the mid-point of RQ and the point R is 3\(\vec {a}\) + 5\(\vec {b}\).

Question 10.

Two adjacent sides of a parallelogram are 2\(\hat {i} \) – 4\(\hat {j} \) + 5\(\hat {k} \) and \(\hat {i} \) – 2\(\hat {j} \) – 3\(\hat {k} \).

Find the unit vector parallel to its diagonal. Also, find its area.

Solution:

The adjacent sides of a parallelogram are \(\vec {a}\) = 2\(\hat {i} \) – 4\(\hat {j} \) + 5\(\hat {k} \) and \(\vec {b}\) = \(\hat {i} \) – 2\(\hat {j} \) – 3\(\hat {k} \).

∴ Its diagonal is \(\vec {a}\) + \(\vec {b}\) = (2\(\hat {i} \) – 4\(\hat {j} \) + 5\(\hat {k} \)) + (\(\hat {i} \) – 2\(\hat {j} \) – 3\(\hat {k} \))

= 3\(\hat {i} \) – 6\(\hat {j} \) + 2\(\hat {k} \).

|\(\vec {a}\) + \(\vec {b}\)| = |3\(\hat {i} \) – 6\(\hat {j} \) + 2\(\hat {k} \)| = \(\sqrt{3^{2}+(-6)^{2}+2^{2}}\)

= \(\sqrt{9+36+4}\) = 7.

∴ Unit vector parallel to diagonal of the parallelogram is \(\frac{1}{7}\)(3\(\hat {i} \) – 6\(\hat {j} \) + 2\(\hat {k} \)) or \(\frac{3}{7}\)\(\hat {i} \) – \(\frac{6}{7}\)\(\hat {j} \) + \(\frac{2}{7}\)\(\hat {k} \).

Thus, the unit vector parallel to its diagonal is \(\frac{1}{7}\)(3\(\hat {i} \) – 6\(\hat {j} \) + 2\(\hat {k} \)) and area of || gm is 11\(\sqrt{5}\) sq.units.

Question 11.

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) and \(\frac{1}{\sqrt{3}}\) respectively.

Solution:

Let a_{1}, a_{2}, a_{3} are the direction ratios of the vector.

Now, vector is equally inclined to the axes.

∴ a_{1} = a_{2} = a_{3} = p (say)

⇒ \(\vec {a}\) = p\(\hat {i} \) + p\(\hat {j} \) + p\(\hat {k} \)

Hence, \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) are the direction cosines of the line equally inclined to the axes.

Question 12.

Let \(\vec {a}\) = \(\hat {i} \) + 4\(\hat {j} \) + 2\(\hat {k} \), \(\vec {b}\) = 3\(\hat {i} \) – 2\(\hat {j} \) + 7\(\hat {k} \) and \(\vec {c}\) = 2\(\hat {i} \) – \(\hat {j} \) + 4\(\hat {k} \).

Find a vector \(\vec {d}\) which is perpendicular to both \(\vec {a}\) and \(\vec {b}\) and \(\vec {c}\).\(\vec {d}\) = 15.

Solution:

Let the vector \(\vec {d}\) be d_{1}\(\hat {i} \) + d_{2}\(\hat {j} \) + d_{3}\(\hat {k} \).

Vector \(\vec {d}\) is perpendicular to vector \(\vec {a}\) = \(\hat {i} \) + 4\(\hat {j} \) + 2\(\hat {k} \)

⇒ d_{1} + 4d_{2} + 2d_{3} = 0 ……………… (1)

Again vector \(\vec {d}\) is perpendicular to vector \(\vec {b}\) = 3\(\hat {i} \) – 2\(\hat {j} \) + 7\(\hat {k} \)

∴ 3d_{1} – 2d_{2} + 7d_{3} = 0 ……………….. (2)

From (1) and (2),

Question 13.

The scalar product of vector \(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \) with a unit vector along the sum of vectors

2\(\hat {i} \) + 4\(\hat {j} \) – 5\(\hat {k} \) and λ\(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \) is equal to 1. Find the value of λ.

Solution:

Sum of the vectors 2\(\hat {i} \) + 4\(\hat {j} \) – 5\(\hat {k} \) and λ\(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \)

= (2\(\hat {i} \) + 4\(\hat {j} \) – 5\(\hat {k} \)) + (λ\(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \))

= (2 + λ)\(\hat {i} \) + 6\(\hat {j} \) – 2\(\hat {k} \) = \(\vec {s}\) (say)

⇒ λ + 6 = r

⇒ (λ + 6)^{2} = r^{2} = (λ^{2} + 4λ + 44) [From (1)]

⇒ λ^{2} + 12λ + 36 = λ^{2} + 4λ + 44

⇒ 8λ = 44 – 36 = 8

⇒ λ = 1.

Question 14.

If \(\vec {a}\), \(\vec {b}\) and \(\vec {c}\) are mutually perpendicular vectors of equal magnitude, show that the vectors \(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\) is equally inclined to \(\vec {a}\), \(\vec {b}\) and \(\vec {c}\).

Solution:

|\(\vec {a}\)| = |\(\vec {b}\)| = |\(\vec {c}\)| = λ(say) ……………… (1)

Also, \(\vec {a}\), \(\vec {b}\) and \(\vec {c}\) are mutually perpendicular.

∴ \(\vec {a}\).\(\vec {b}\) = 0, \(\vec {b}\).\(\vec {c}\) = 0 and \(\vec {a}\) + \(\vec {c}\) = 0 ……………… (2)

Let θ be the angle between vectors \(\vec {a}\) and \(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\).

Question 15.

Prove that (\(\vec {a}\) + \(\vec {b}\)).(\(\vec {a}\) + \(\vec {b}\)) = |\(\vec {a}\)|^{2} + |\(\vec {b}\)|^{2},

if any and only if \(\vec {a}\) and \(\vec {b}\) are perpendicular to each other, given that \(\vec {a}\) ≠ 0, \(\vec {b}\) ≠ 0.

Solution:

(\(\vec {a}\) + \(\vec {b}\)).(\(\vec {a}\) + \(\vec {b}\)) = \(\vec {a}\).(\(\vec {a}\) + \(\vec {b}\)) + \(\vec {b}\).(\(\vec {a}\) + \(\vec {b}\))

= \(\vec {a}\).\(\vec {a}\) + \(\vec {a}\).\(\vec {b}\) + \(\vec {b}\).\(\vec {a}\) + \(\vec {b}\).\(\vec {b}\)

= |\(\vec {a}\)|^{2} + \(\vec {a}\).\(\vec {b}\) + \(\vec {b}\).\(\vec {a}\) + |\(\vec {b}\)|^{2}

= |\(\vec {a}\)|^{2} + |\(\vec {b}\)|^{2} + 2\(\vec {a}\)\(\vec {b}\) [∵ \(\vec {b}\).\(\vec {a}\) = \(\vec {a}\).\(\vec {b}\)

(i) Let \(\vec {a}\) ⊥ \(\vec {b}\). So, \(\vec {a}\).\(\vec {b}\) = 0.

∴ (\(\vec {a}\) + \(\vec {b}\)).(\(\vec {a}\) + \(\vec {b}\)) = |\(\vec {a}\)|^{2} + |\(\vec {b}\)|^{2} + 2\(\vec {a}\) \(\vec {b}\)

= |\(\vec {a}\)|^{2} + |\(\vec {b}\)|^{2} [∵ \(\vec {a}\).\(\vec {b}\) = 0]

Again if \(\vec {a}\) is not perpendicular to \(\vec {b}\) ⇒ \(\vec {a}\).\(\vec {b}\) ≠ 0.

Now, (\(\vec {a}\) + \(\vec {b}\)).(\(\vec {a}\) + \(\vec {b}\)) = |\(\vec {a}\)|^{2} + |\(\vec {b}\)|^{2} + 2 \(\vec {a}\) \(\vec {b}\)

≠ |\(\vec {a}\)|^{2} + |\(\vec {b}\)|^{2}

unless \(\vec {a}\).\(\vec {b}\) = 0.

Hence, \(\vec {a}\).\(\vec {b}\) = 0 or vector \(\vec {a}\) is perpendicular to \(\vec {b}\).

Choose the correct answers in questions 16 to 19:

Question 16.

If θ is the angle between two vectors \(\vec {a}\) and \(\vec {b}\), then \(\vec {a}\), \(\vec {b}\) ≥ 0 only, when

(A) 0 < θ < \(\frac{π}{2}\)

(B) θ = \(\frac{π}{3}\)

(C) θ = \(\frac{π}{2}\)

(D) 0 ≤ θ ≤ π

Solution:

\(\vec {a}\).\(\vec {b}\) ≥ 0 ⇒ |\(\vec {a}\)| |\(\vec {b}\)| cos θ ≥ 0

Now, |\(\vec {a}\)| = +ve, |\(\vec {b}\)| = +ve ⇒ cos θ ≥ 0

Therefore, 0 ≤ θ ≤ \(\frac{π}{2}\)

∴ Part (B) is the correct answer.

Question 17.

Let \(\vec {a}\) and \(\vec {b}\) be two unit vectors and θ be the angle between them. Then, \(\vec {a}\) + \(\vec {b}\) is a unit vector, if

(A) θ = \(\frac{π}{4}\)

(B) θ = \(\frac{π}{3}\)

(C) θ = \(\frac{π}{2}\)

(D) θ = \(\frac{2π}{3}\)

Solution:

We have:

|\(\vec {a}\)| = 1, |\(\vec {b}\)| = 1

and |\(\vec {a}\) + \(\vec {b}\)| = 1, or |\(\vec {a}\) + \(\vec {b}\)|^{2} = 1.

i.e. (\(\vec {a}\) + \(\vec {b}\)). (\(\vec {a}\) + \(\vec {b}\)) = 1

or |\(\vec {a}\)|^{2} + |\(\vec {b}\)|^{2} + 2\(\vec {a}\)\(\vec {b}\) = 1 [ \(\vec {b}\).\(\vec {a}\) = \(\vec {a}\).\(\vec {b}\)]

= 1 + 1 + 2|\(\vec {a}\)| |\(\vec {b}\)| cos θ = 1

[∵ |\(\vec {a}\)| = 1, |\(\vec {b}\)| = 1],

where θ is the angle between \(\vec {a}\) and \(\vec {b}\).

∴ 2 + 2 . 1 . cos θ = 1

or 2 cos θ = 1 ⇒ cos θ = – \(\frac{1}{2}\)

⇒ θ = \(\frac{2π}{3}\).

∴ Part (D) is the correct answer.

Question 18.

The value of \(\hat {i} \).(\(\hat {j} \) × \(\hat {k} \)) + \(\hat {j} \).(\(\hat {i} \) × \(\hat {j} \)) is

(A) 0

(B) \(\frac{π}{4}\)

(C) \(\frac{π}{2}\)

(D) π

Solution:

(\(\hat {i} \) + \(\hat {j} \) + \(\hat {k} \)) are the perpendicular unit vectors, we know that

\(\hat {i} \) × \(\hat {j} \) = \(\hat {k} \), \(\hat {j} \) × \(\hat {k} \) = \(\hat {i} \), \(\hat {k} \) × \(\hat {i} \) = \(\hat {j} \)

and \(\hat {i} \).\(\hat {i} \) = 1, \(\hat {j} \).\(\hat {j} \) = 1 and \(\hat {k} \).\(\hat {k} \) = 1

⇒ \(\hat {i} \).(\(\hat {j} \) × \(\hat {k} \)) + \(\hat {j} \).(\(\hat {k} \) × \(\hat {i} \)) + \(\hat {k} \).(\(\hat {i} \) × \(\hat {j} \))

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

[∵ \(\hat {i} \) × \(\hat {k} \) = – \(\hat {k} \) × \(\hat {i} \)]

= \(\hat {i} \).\(\hat {i} \) – \(\hat {j} \).\(\hat {j} \) + \(\hat {k} \).\(\hat {k} \) = 1 – 1 + 1 = 1.

∴ Part (C) is the correct answer.

Question 19.

If θ is the angle between any two vectors \(\vec {a}\) and \(\vec {b}\), then |\(\vec {a}\).\(\vec {b}\)| = |\(\vec {a}\) × \(\vec {b}\)|, when θ is equal to

(A) 0

(B) \(\frac{π}{4}\)

(C) \(\frac{π}{2}\)

(D) π

Solution:

θ is angle between vectors \(\vec {a}\) and \(\vec {b}\)

∴ |\(\vec {a}\).\(\vec {b}\)| = |\(\vec {a}\)| |\(\vec {b}\)| |cos θ|

and |\(\vec {a}\) × \(\vec {b}\)| = |\(\vec {a}\)| |\(\vec {b}\)| |sin θ|

We have: |\(\vec {a}\).\(\vec {b}\)| = |\(\vec {a}\) × \(\vec {b}\)|

∴ |\(\vec {a}\)| |\(\vec {b}\)| |cos θ| = |\(\vec {a}\)| |\(\vec {b}\)| |sin θ|

⇒ |cos θ| = |sin θ|

or |tan θ| = 1 or tan θ = 1

⇒ θ = \(\frac{π}{4}\).

∴ Part (B) is the correct answer.