QBManager

Vector Algebra

87680 For 3 points $A (\vec{a}), B (\vec{b}), C (\vec{c})$
if $3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$ , then

1 Point

C

divides

AB

externally in ratio

3 : 2

2 3 Points from

△ ABC

3

C

is not mid-point of

A B

4

C

divides

A B

internally in ratio

2 : 3

Explanation:

(D)Given that,
$3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$
$5 \vec{c} = 3 \vec{a} + 2 \vec{b}$
$\vec{c} = \frac{3 \vec{a} + 2 \vec{b}}{5} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$\vec{c} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$c = \frac{3 (\frac{2}{3} \vec{b} + \vec{a})}{3 (\frac{2}{3} + 1)}$
$c = \frac{a + λ b}{λ + 1}$
Comparing equation (i) \& (ii), we get-
$λ = \frac{2}{3}$
This shows that the point $C$ divides segment $A B$ internally in the ratio $2 : 3$ .

MHT CET-2007

Vector Algebra

87660 If $\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
and $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$ are the position vectors of the vertices of triangle $A B C$ respectively, then the position vector of the intersection of the medians of the triangle $A B C$ is

1

2 \hat{i} + 3 \hat{j} + 3 \hat{k}

2

4 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

5 \hat{i} + 3 \hat{j} + 3 \hat{k}

4

3 \hat{i} + 3 \hat{j} + 4 \hat{k}

Explanation:

(B) : Given, Position vector,
$\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
$\vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
And, $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$
original image
Intersection median of triangle $A^{6} B^{1} C^{5}$ is $G$ which is centroid of triangle.
Centroid of triangle $(G) = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$= \frac{2 \hat{i} + 3 \hat{j} + \hat{k} + 4 \hat{i} + 5 \hat{j} + 3 \hat{k} + 6 \hat{i} + \hat{j} + 5 \hat{k}}{3}$
$= \frac{12 \hat{i} + 9 \hat{j} + 9 \hat{k}}{3} = 4 \hat{i} + 3 \hat{j} + 3 \hat{k}$

MHT CET-2020

Vector Algebra

87679 Let $A (3, 5, 6)$ and $B (4, 6, - 3)$ . Find ratio in which yz plane is dividing $A B$ .

1

3 : 4

externally

2

3 : 4

internally

3

4 : 3

externally

4

4 : 3

internally

Explanation:

(A) : Given,
$A = (3, 5, 6)$ and $B = (4, 6, - 3)$
Let, $y z$ -plane divide $A B$ in the ratio of $k : 1$ at point $P$ . $x$ coordinate of point $P$ will be 0 .
$(x_{1} = 3, x_{2} = 4)$
$P = \frac{{kx}_{2} + x_{1}}{k + 1}$ ,
$0 = \frac{k (4) + 1 (3)}{k + 1}$
$4 k + 3 = 0$
$\frac{k}{1} = \frac{- 3}{4}$
$k : 1 = - 3 : 4$
$∵$ Externally ratio is negative.
Hence, YZ-plane divides the segment AB externally in the ratio of $3 : 4$ .

MHT CET-2008

Vector Algebra

87661 If the position vectors of the vertices, $A, B, C$ of a triangle $A B C$ are $4 \hat{i} + 7 \hat{j} + 8 \hat{k}, 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ respectively, then the position vector of the point where bisector of angel $A$ meets $B C$ is

1

\frac{1}{3} (6 \hat{i} + 11 \hat{j} + 15 \hat{k})

2

\frac{1}{4} (8 \hat{i} + 14 \hat{j} + 19 \hat{k})

3

\frac{1}{2} (4 \hat{i} + 8 \hat{j} + 11 \hat{k})

4

\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})

Explanation:

(D) : We have, The position vector,
$\vec{a} = 4 \hat{i} + 7 \hat{j} + 8 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}$
original image
$\vec{AB} = (\vec{b} - \vec{a}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$= - 2 \hat{i} - 4 \hat{j} - 4 \hat{k}$
$| \vec{AB} | = \sqrt{4 + 16 + 16} = 6$
$\vec{AC} = (\vec{c} - \vec{a})$
$\vec{AC} = (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$\vec{AC} = - 2 \hat{i} - 2 \hat{j} - \hat{k}$
$| \vec{AC} | = \sqrt{4 + 4 + 1} = 3$
Then $A B$ divides $B C$ in the ratio $A B : A C$
Position vector, Of D
$= \frac{| \vec{AB} | (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + | \vec{AC} | (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{| \vec{AB} + | \vec{AC} ∣}$
$= \frac{6 (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + 3 (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{6 + 3}$
$= \frac{(12 \hat{i} + 30 \hat{j} + 42 \hat{k} + 6 \hat{i} + 9 \hat{j} + 12 \hat{k})}{6 + 3}$
$= \frac{(18 \hat{i} + 39 \hat{j} + 54 \hat{k})}{6 + 3} = \frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$

MHT CET-2020

Vector Algebra

87662 In a quadrilateral $ABCD, M$ and $N$ are the midpoints of the sides $A B$ and $C D$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$ , then $t =$

1

\frac{1}{2}

2

\frac{3}{2}

3 2

4 4

Explanation:

(C) : Let $\vec{a}, \vec{b}, \vec{c} \vec{d}, \vec{m}, \vec{n}$ be the position vectors of A, B, C, D, M, N, respectively.
$M$ and $N$ are the midpoints of $A B$ and $C D$ respectively.
original image
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$ ,
$\vec{a} + \vec{b} = 2 \vec{m}$
$\vec{n} = \frac{\vec{c} + \vec{d}}{2}$
$\vec{c} + \vec{d} = 2 \vec{n}$
$\vec{AD} = (\vec{d} - \vec{a}), \vec{BC} = (\vec{c} - \vec{b})$
Given,
$\vec{AD} + \vec{BC} = t (\overset{―}{MN})$
$(\vec{d} - \vec{a}) + (\vec{c} - \vec{b}) = t (\vec{n} - \vec{m})$
$\vec{d} - \vec{a} + \vec{c} - \vec{b} = t (\vec{n} - \vec{m})$
$(\vec{d} + \vec{c}) - (\vec{a} + \vec{b}) = t (\vec{n} - \vec{m})$
$2 \vec{n} - 2 \vec{m} = t (\vec{n} - \vec{m})$
$2 (\vec{n} - \vec{m}) = t (\vec{n} - \vec{m})$
So, $t = 2$

MHT CET-2020

Vector Algebra

87680 For 3 points $A (\vec{a}), B (\vec{b}), C (\vec{c})$
if $3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$ , then

1 Point

C

divides

AB

externally in ratio

3 : 2

2 3 Points from

△ ABC

3

C

is not mid-point of

A B

4

C

divides

A B

internally in ratio

2 : 3

Explanation:

(D)Given that,
$3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$
$5 \vec{c} = 3 \vec{a} + 2 \vec{b}$
$\vec{c} = \frac{3 \vec{a} + 2 \vec{b}}{5} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$\vec{c} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$c = \frac{3 (\frac{2}{3} \vec{b} + \vec{a})}{3 (\frac{2}{3} + 1)}$
$c = \frac{a + λ b}{λ + 1}$
Comparing equation (i) \& (ii), we get-
$λ = \frac{2}{3}$
This shows that the point $C$ divides segment $A B$ internally in the ratio $2 : 3$ .

MHT CET-2007

Vector Algebra

87660 If $\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
and $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$ are the position vectors of the vertices of triangle $A B C$ respectively, then the position vector of the intersection of the medians of the triangle $A B C$ is

1

2 \hat{i} + 3 \hat{j} + 3 \hat{k}

2

4 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

5 \hat{i} + 3 \hat{j} + 3 \hat{k}

4

3 \hat{i} + 3 \hat{j} + 4 \hat{k}

Explanation:

(B) : Given, Position vector,
$\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
$\vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
And, $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$
original image
Intersection median of triangle $A^{6} B^{1} C^{5}$ is $G$ which is centroid of triangle.
Centroid of triangle $(G) = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$= \frac{2 \hat{i} + 3 \hat{j} + \hat{k} + 4 \hat{i} + 5 \hat{j} + 3 \hat{k} + 6 \hat{i} + \hat{j} + 5 \hat{k}}{3}$
$= \frac{12 \hat{i} + 9 \hat{j} + 9 \hat{k}}{3} = 4 \hat{i} + 3 \hat{j} + 3 \hat{k}$

MHT CET-2020

Vector Algebra

87679 Let $A (3, 5, 6)$ and $B (4, 6, - 3)$ . Find ratio in which yz plane is dividing $A B$ .

1

3 : 4

externally

2

3 : 4

internally

3

4 : 3

externally

4

4 : 3

internally

Explanation:

(A) : Given,
$A = (3, 5, 6)$ and $B = (4, 6, - 3)$
Let, $y z$ -plane divide $A B$ in the ratio of $k : 1$ at point $P$ . $x$ coordinate of point $P$ will be 0 .
$(x_{1} = 3, x_{2} = 4)$
$P = \frac{{kx}_{2} + x_{1}}{k + 1}$ ,
$0 = \frac{k (4) + 1 (3)}{k + 1}$
$4 k + 3 = 0$
$\frac{k}{1} = \frac{- 3}{4}$
$k : 1 = - 3 : 4$
$∵$ Externally ratio is negative.
Hence, YZ-plane divides the segment AB externally in the ratio of $3 : 4$ .

MHT CET-2008

Vector Algebra

87661 If the position vectors of the vertices, $A, B, C$ of a triangle $A B C$ are $4 \hat{i} + 7 \hat{j} + 8 \hat{k}, 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ respectively, then the position vector of the point where bisector of angel $A$ meets $B C$ is

1

\frac{1}{3} (6 \hat{i} + 11 \hat{j} + 15 \hat{k})

2

\frac{1}{4} (8 \hat{i} + 14 \hat{j} + 19 \hat{k})

3

\frac{1}{2} (4 \hat{i} + 8 \hat{j} + 11 \hat{k})

4

\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})

Explanation:

(D) : We have, The position vector,
$\vec{a} = 4 \hat{i} + 7 \hat{j} + 8 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}$
original image
$\vec{AB} = (\vec{b} - \vec{a}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$= - 2 \hat{i} - 4 \hat{j} - 4 \hat{k}$
$| \vec{AB} | = \sqrt{4 + 16 + 16} = 6$
$\vec{AC} = (\vec{c} - \vec{a})$
$\vec{AC} = (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$\vec{AC} = - 2 \hat{i} - 2 \hat{j} - \hat{k}$
$| \vec{AC} | = \sqrt{4 + 4 + 1} = 3$
Then $A B$ divides $B C$ in the ratio $A B : A C$
Position vector, Of D
$= \frac{| \vec{AB} | (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + | \vec{AC} | (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{| \vec{AB} + | \vec{AC} ∣}$
$= \frac{6 (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + 3 (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{6 + 3}$
$= \frac{(12 \hat{i} + 30 \hat{j} + 42 \hat{k} + 6 \hat{i} + 9 \hat{j} + 12 \hat{k})}{6 + 3}$
$= \frac{(18 \hat{i} + 39 \hat{j} + 54 \hat{k})}{6 + 3} = \frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$

MHT CET-2020

Vector Algebra

87662 In a quadrilateral $ABCD, M$ and $N$ are the midpoints of the sides $A B$ and $C D$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$ , then $t =$

1

\frac{1}{2}

2

\frac{3}{2}

3 2

4 4

Explanation:

(C) : Let $\vec{a}, \vec{b}, \vec{c} \vec{d}, \vec{m}, \vec{n}$ be the position vectors of A, B, C, D, M, N, respectively.
$M$ and $N$ are the midpoints of $A B$ and $C D$ respectively.
original image
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$ ,
$\vec{a} + \vec{b} = 2 \vec{m}$
$\vec{n} = \frac{\vec{c} + \vec{d}}{2}$
$\vec{c} + \vec{d} = 2 \vec{n}$
$\vec{AD} = (\vec{d} - \vec{a}), \vec{BC} = (\vec{c} - \vec{b})$
Given,
$\vec{AD} + \vec{BC} = t (\overset{―}{MN})$
$(\vec{d} - \vec{a}) + (\vec{c} - \vec{b}) = t (\vec{n} - \vec{m})$
$\vec{d} - \vec{a} + \vec{c} - \vec{b} = t (\vec{n} - \vec{m})$
$(\vec{d} + \vec{c}) - (\vec{a} + \vec{b}) = t (\vec{n} - \vec{m})$
$2 \vec{n} - 2 \vec{m} = t (\vec{n} - \vec{m})$
$2 (\vec{n} - \vec{m}) = t (\vec{n} - \vec{m})$
So, $t = 2$

MHT CET-2020

Vector Algebra

87680 For 3 points $A (\vec{a}), B (\vec{b}), C (\vec{c})$
if $3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$ , then

1 Point

C

divides

AB

externally in ratio

3 : 2

2 3 Points from

△ ABC

3

C

is not mid-point of

A B

4

C

divides

A B

internally in ratio

2 : 3

Explanation:

(D)Given that,
$3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$
$5 \vec{c} = 3 \vec{a} + 2 \vec{b}$
$\vec{c} = \frac{3 \vec{a} + 2 \vec{b}}{5} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$\vec{c} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$c = \frac{3 (\frac{2}{3} \vec{b} + \vec{a})}{3 (\frac{2}{3} + 1)}$
$c = \frac{a + λ b}{λ + 1}$
Comparing equation (i) \& (ii), we get-
$λ = \frac{2}{3}$
This shows that the point $C$ divides segment $A B$ internally in the ratio $2 : 3$ .

MHT CET-2007

Vector Algebra

87660 If $\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
and $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$ are the position vectors of the vertices of triangle $A B C$ respectively, then the position vector of the intersection of the medians of the triangle $A B C$ is

1

2 \hat{i} + 3 \hat{j} + 3 \hat{k}

2

4 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

5 \hat{i} + 3 \hat{j} + 3 \hat{k}

4

3 \hat{i} + 3 \hat{j} + 4 \hat{k}

Explanation:

(B) : Given, Position vector,
$\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
$\vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
And, $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$
original image
Intersection median of triangle $A^{6} B^{1} C^{5}$ is $G$ which is centroid of triangle.
Centroid of triangle $(G) = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$= \frac{2 \hat{i} + 3 \hat{j} + \hat{k} + 4 \hat{i} + 5 \hat{j} + 3 \hat{k} + 6 \hat{i} + \hat{j} + 5 \hat{k}}{3}$
$= \frac{12 \hat{i} + 9 \hat{j} + 9 \hat{k}}{3} = 4 \hat{i} + 3 \hat{j} + 3 \hat{k}$

MHT CET-2020

Vector Algebra

87679 Let $A (3, 5, 6)$ and $B (4, 6, - 3)$ . Find ratio in which yz plane is dividing $A B$ .

1

3 : 4

externally

2

3 : 4

internally

3

4 : 3

externally

4

4 : 3

internally

Explanation:

(A) : Given,
$A = (3, 5, 6)$ and $B = (4, 6, - 3)$
Let, $y z$ -plane divide $A B$ in the ratio of $k : 1$ at point $P$ . $x$ coordinate of point $P$ will be 0 .
$(x_{1} = 3, x_{2} = 4)$
$P = \frac{{kx}_{2} + x_{1}}{k + 1}$ ,
$0 = \frac{k (4) + 1 (3)}{k + 1}$
$4 k + 3 = 0$
$\frac{k}{1} = \frac{- 3}{4}$
$k : 1 = - 3 : 4$
$∵$ Externally ratio is negative.
Hence, YZ-plane divides the segment AB externally in the ratio of $3 : 4$ .

MHT CET-2008

Vector Algebra

87661 If the position vectors of the vertices, $A, B, C$ of a triangle $A B C$ are $4 \hat{i} + 7 \hat{j} + 8 \hat{k}, 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ respectively, then the position vector of the point where bisector of angel $A$ meets $B C$ is

1

\frac{1}{3} (6 \hat{i} + 11 \hat{j} + 15 \hat{k})

2

\frac{1}{4} (8 \hat{i} + 14 \hat{j} + 19 \hat{k})

3

\frac{1}{2} (4 \hat{i} + 8 \hat{j} + 11 \hat{k})

4

\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})

Explanation:

(D) : We have, The position vector,
$\vec{a} = 4 \hat{i} + 7 \hat{j} + 8 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}$
original image
$\vec{AB} = (\vec{b} - \vec{a}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$= - 2 \hat{i} - 4 \hat{j} - 4 \hat{k}$
$| \vec{AB} | = \sqrt{4 + 16 + 16} = 6$
$\vec{AC} = (\vec{c} - \vec{a})$
$\vec{AC} = (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$\vec{AC} = - 2 \hat{i} - 2 \hat{j} - \hat{k}$
$| \vec{AC} | = \sqrt{4 + 4 + 1} = 3$
Then $A B$ divides $B C$ in the ratio $A B : A C$
Position vector, Of D
$= \frac{| \vec{AB} | (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + | \vec{AC} | (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{| \vec{AB} + | \vec{AC} ∣}$
$= \frac{6 (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + 3 (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{6 + 3}$
$= \frac{(12 \hat{i} + 30 \hat{j} + 42 \hat{k} + 6 \hat{i} + 9 \hat{j} + 12 \hat{k})}{6 + 3}$
$= \frac{(18 \hat{i} + 39 \hat{j} + 54 \hat{k})}{6 + 3} = \frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$

MHT CET-2020

Vector Algebra

87662 In a quadrilateral $ABCD, M$ and $N$ are the midpoints of the sides $A B$ and $C D$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$ , then $t =$

1

\frac{1}{2}

2

\frac{3}{2}

3 2

4 4

Explanation:

(C) : Let $\vec{a}, \vec{b}, \vec{c} \vec{d}, \vec{m}, \vec{n}$ be the position vectors of A, B, C, D, M, N, respectively.
$M$ and $N$ are the midpoints of $A B$ and $C D$ respectively.
original image
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$ ,
$\vec{a} + \vec{b} = 2 \vec{m}$
$\vec{n} = \frac{\vec{c} + \vec{d}}{2}$
$\vec{c} + \vec{d} = 2 \vec{n}$
$\vec{AD} = (\vec{d} - \vec{a}), \vec{BC} = (\vec{c} - \vec{b})$
Given,
$\vec{AD} + \vec{BC} = t (\overset{―}{MN})$
$(\vec{d} - \vec{a}) + (\vec{c} - \vec{b}) = t (\vec{n} - \vec{m})$
$\vec{d} - \vec{a} + \vec{c} - \vec{b} = t (\vec{n} - \vec{m})$
$(\vec{d} + \vec{c}) - (\vec{a} + \vec{b}) = t (\vec{n} - \vec{m})$
$2 \vec{n} - 2 \vec{m} = t (\vec{n} - \vec{m})$
$2 (\vec{n} - \vec{m}) = t (\vec{n} - \vec{m})$
So, $t = 2$

MHT CET-2020

Vector Algebra

87680 For 3 points $A (\vec{a}), B (\vec{b}), C (\vec{c})$
if $3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$ , then

1 Point

C

divides

AB

externally in ratio

3 : 2

2 3 Points from

△ ABC

3

C

is not mid-point of

A B

4

C

divides

A B

internally in ratio

2 : 3

Explanation:

(D)Given that,
$3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$
$5 \vec{c} = 3 \vec{a} + 2 \vec{b}$
$\vec{c} = \frac{3 \vec{a} + 2 \vec{b}}{5} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$\vec{c} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$c = \frac{3 (\frac{2}{3} \vec{b} + \vec{a})}{3 (\frac{2}{3} + 1)}$
$c = \frac{a + λ b}{λ + 1}$
Comparing equation (i) \& (ii), we get-
$λ = \frac{2}{3}$
This shows that the point $C$ divides segment $A B$ internally in the ratio $2 : 3$ .

MHT CET-2007

Vector Algebra

87660 If $\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
and $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$ are the position vectors of the vertices of triangle $A B C$ respectively, then the position vector of the intersection of the medians of the triangle $A B C$ is

1

2 \hat{i} + 3 \hat{j} + 3 \hat{k}

2

4 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

5 \hat{i} + 3 \hat{j} + 3 \hat{k}

4

3 \hat{i} + 3 \hat{j} + 4 \hat{k}

Explanation:

(B) : Given, Position vector,
$\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
$\vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
And, $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$
original image
Intersection median of triangle $A^{6} B^{1} C^{5}$ is $G$ which is centroid of triangle.
Centroid of triangle $(G) = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$= \frac{2 \hat{i} + 3 \hat{j} + \hat{k} + 4 \hat{i} + 5 \hat{j} + 3 \hat{k} + 6 \hat{i} + \hat{j} + 5 \hat{k}}{3}$
$= \frac{12 \hat{i} + 9 \hat{j} + 9 \hat{k}}{3} = 4 \hat{i} + 3 \hat{j} + 3 \hat{k}$

MHT CET-2020

Vector Algebra

87679 Let $A (3, 5, 6)$ and $B (4, 6, - 3)$ . Find ratio in which yz plane is dividing $A B$ .

1

3 : 4

externally

2

3 : 4

internally

3

4 : 3

externally

4

4 : 3

internally

Explanation:

(A) : Given,
$A = (3, 5, 6)$ and $B = (4, 6, - 3)$
Let, $y z$ -plane divide $A B$ in the ratio of $k : 1$ at point $P$ . $x$ coordinate of point $P$ will be 0 .
$(x_{1} = 3, x_{2} = 4)$
$P = \frac{{kx}_{2} + x_{1}}{k + 1}$ ,
$0 = \frac{k (4) + 1 (3)}{k + 1}$
$4 k + 3 = 0$
$\frac{k}{1} = \frac{- 3}{4}$
$k : 1 = - 3 : 4$
$∵$ Externally ratio is negative.
Hence, YZ-plane divides the segment AB externally in the ratio of $3 : 4$ .

MHT CET-2008

Vector Algebra

87661 If the position vectors of the vertices, $A, B, C$ of a triangle $A B C$ are $4 \hat{i} + 7 \hat{j} + 8 \hat{k}, 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ respectively, then the position vector of the point where bisector of angel $A$ meets $B C$ is

1

\frac{1}{3} (6 \hat{i} + 11 \hat{j} + 15 \hat{k})

2

\frac{1}{4} (8 \hat{i} + 14 \hat{j} + 19 \hat{k})

3

\frac{1}{2} (4 \hat{i} + 8 \hat{j} + 11 \hat{k})

4

\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})

Explanation:

(D) : We have, The position vector,
$\vec{a} = 4 \hat{i} + 7 \hat{j} + 8 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}$
original image
$\vec{AB} = (\vec{b} - \vec{a}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$= - 2 \hat{i} - 4 \hat{j} - 4 \hat{k}$
$| \vec{AB} | = \sqrt{4 + 16 + 16} = 6$
$\vec{AC} = (\vec{c} - \vec{a})$
$\vec{AC} = (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$\vec{AC} = - 2 \hat{i} - 2 \hat{j} - \hat{k}$
$| \vec{AC} | = \sqrt{4 + 4 + 1} = 3$
Then $A B$ divides $B C$ in the ratio $A B : A C$
Position vector, Of D
$= \frac{| \vec{AB} | (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + | \vec{AC} | (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{| \vec{AB} + | \vec{AC} ∣}$
$= \frac{6 (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + 3 (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{6 + 3}$
$= \frac{(12 \hat{i} + 30 \hat{j} + 42 \hat{k} + 6 \hat{i} + 9 \hat{j} + 12 \hat{k})}{6 + 3}$
$= \frac{(18 \hat{i} + 39 \hat{j} + 54 \hat{k})}{6 + 3} = \frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$

MHT CET-2020

Vector Algebra

87662 In a quadrilateral $ABCD, M$ and $N$ are the midpoints of the sides $A B$ and $C D$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$ , then $t =$

1

\frac{1}{2}

2

\frac{3}{2}

3 2

4 4

Explanation:

(C) : Let $\vec{a}, \vec{b}, \vec{c} \vec{d}, \vec{m}, \vec{n}$ be the position vectors of A, B, C, D, M, N, respectively.
$M$ and $N$ are the midpoints of $A B$ and $C D$ respectively.
original image
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$ ,
$\vec{a} + \vec{b} = 2 \vec{m}$
$\vec{n} = \frac{\vec{c} + \vec{d}}{2}$
$\vec{c} + \vec{d} = 2 \vec{n}$
$\vec{AD} = (\vec{d} - \vec{a}), \vec{BC} = (\vec{c} - \vec{b})$
Given,
$\vec{AD} + \vec{BC} = t (\overset{―}{MN})$
$(\vec{d} - \vec{a}) + (\vec{c} - \vec{b}) = t (\vec{n} - \vec{m})$
$\vec{d} - \vec{a} + \vec{c} - \vec{b} = t (\vec{n} - \vec{m})$
$(\vec{d} + \vec{c}) - (\vec{a} + \vec{b}) = t (\vec{n} - \vec{m})$
$2 \vec{n} - 2 \vec{m} = t (\vec{n} - \vec{m})$
$2 (\vec{n} - \vec{m}) = t (\vec{n} - \vec{m})$
So, $t = 2$

MHT CET-2020

Vector Algebra

87680 For 3 points $A (\vec{a}), B (\vec{b}), C (\vec{c})$
if $3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$ , then

1 Point

C

divides

AB

externally in ratio

3 : 2

2 3 Points from

△ ABC

3

C

is not mid-point of

A B

4

C

divides

A B

internally in ratio

2 : 3

Explanation:

(D)Given that,
$3 \vec{a} + 2 \vec{b} - 5 \vec{c} = 0$
$5 \vec{c} = 3 \vec{a} + 2 \vec{b}$
$\vec{c} = \frac{3 \vec{a} + 2 \vec{b}}{5} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$\vec{c} = \frac{2 \vec{b} + 3 \vec{a}}{2 + 3}$
$c = \frac{3 (\frac{2}{3} \vec{b} + \vec{a})}{3 (\frac{2}{3} + 1)}$
$c = \frac{a + λ b}{λ + 1}$
Comparing equation (i) \& (ii), we get-
$λ = \frac{2}{3}$
This shows that the point $C$ divides segment $A B$ internally in the ratio $2 : 3$ .

MHT CET-2007

Vector Algebra

87660 If $\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}, \vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
and $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$ are the position vectors of the vertices of triangle $A B C$ respectively, then the position vector of the intersection of the medians of the triangle $A B C$ is

1

2 \hat{i} + 3 \hat{j} + 3 \hat{k}

2

4 \hat{i} + 3 \hat{j} + 3 \hat{k}

3

5 \hat{i} + 3 \hat{j} + 3 \hat{k}

4

3 \hat{i} + 3 \hat{j} + 4 \hat{k}

Explanation:

(B) : Given, Position vector,
$\vec{a} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
$\vec{b} = 4 \hat{i} + 5 \hat{j} + 3 \hat{k}$
And, $\vec{c} = 6 \hat{i} + \hat{j} + 5 \hat{k}$
original image
Intersection median of triangle $A^{6} B^{1} C^{5}$ is $G$ which is centroid of triangle.
Centroid of triangle $(G) = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$= \frac{2 \hat{i} + 3 \hat{j} + \hat{k} + 4 \hat{i} + 5 \hat{j} + 3 \hat{k} + 6 \hat{i} + \hat{j} + 5 \hat{k}}{3}$
$= \frac{12 \hat{i} + 9 \hat{j} + 9 \hat{k}}{3} = 4 \hat{i} + 3 \hat{j} + 3 \hat{k}$

MHT CET-2020

Vector Algebra

87679 Let $A (3, 5, 6)$ and $B (4, 6, - 3)$ . Find ratio in which yz plane is dividing $A B$ .

1

3 : 4

externally

2

3 : 4

internally

3

4 : 3

externally

4

4 : 3

internally

Explanation:

(A) : Given,
$A = (3, 5, 6)$ and $B = (4, 6, - 3)$
Let, $y z$ -plane divide $A B$ in the ratio of $k : 1$ at point $P$ . $x$ coordinate of point $P$ will be 0 .
$(x_{1} = 3, x_{2} = 4)$
$P = \frac{{kx}_{2} + x_{1}}{k + 1}$ ,
$0 = \frac{k (4) + 1 (3)}{k + 1}$
$4 k + 3 = 0$
$\frac{k}{1} = \frac{- 3}{4}$
$k : 1 = - 3 : 4$
$∵$ Externally ratio is negative.
Hence, YZ-plane divides the segment AB externally in the ratio of $3 : 4$ .

MHT CET-2008

Vector Algebra

87661 If the position vectors of the vertices, $A, B, C$ of a triangle $A B C$ are $4 \hat{i} + 7 \hat{j} + 8 \hat{k}, 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ respectively, then the position vector of the point where bisector of angel $A$ meets $B C$ is

1

\frac{1}{3} (6 \hat{i} + 11 \hat{j} + 15 \hat{k})

2

\frac{1}{4} (8 \hat{i} + 14 \hat{j} + 19 \hat{k})

3

\frac{1}{2} (4 \hat{i} + 8 \hat{j} + 11 \hat{k})

4

\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})

Explanation:

(D) : We have, The position vector,
$\vec{a} = 4 \hat{i} + 7 \hat{j} + 8 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$
$\vec{c} = 2 \hat{i} + 5 \hat{j} + 7 \hat{k}$
original image
$\vec{AB} = (\vec{b} - \vec{a}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$= - 2 \hat{i} - 4 \hat{j} - 4 \hat{k}$
$| \vec{AB} | = \sqrt{4 + 16 + 16} = 6$
$\vec{AC} = (\vec{c} - \vec{a})$
$\vec{AC} = (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) - (4 \hat{i} + 7 \hat{j} + 8 \hat{k})$
$\vec{AC} = - 2 \hat{i} - 2 \hat{j} - \hat{k}$
$| \vec{AC} | = \sqrt{4 + 4 + 1} = 3$
Then $A B$ divides $B C$ in the ratio $A B : A C$
Position vector, Of D
$= \frac{| \vec{AB} | (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + | \vec{AC} | (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{| \vec{AB} + | \vec{AC} ∣}$
$= \frac{6 (2 \hat{i} + 5 \hat{j} + 7 \hat{k}) + 3 (2 \hat{i} + 3 \hat{j} + 4 \hat{k})}{6 + 3}$
$= \frac{(12 \hat{i} + 30 \hat{j} + 42 \hat{k} + 6 \hat{i} + 9 \hat{j} + 12 \hat{k})}{6 + 3}$
$= \frac{(18 \hat{i} + 39 \hat{j} + 54 \hat{k})}{6 + 3} = \frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})$

MHT CET-2020

Vector Algebra

87662 In a quadrilateral $ABCD, M$ and $N$ are the midpoints of the sides $A B$ and $C D$ respectively. If $\vec{AD} + \vec{BC} = t \vec{MN}$ , then $t =$

1

\frac{1}{2}

2

\frac{3}{2}

3 2

4 4

Explanation:

(C) : Let $\vec{a}, \vec{b}, \vec{c} \vec{d}, \vec{m}, \vec{n}$ be the position vectors of A, B, C, D, M, N, respectively.
$M$ and $N$ are the midpoints of $A B$ and $C D$ respectively.
original image
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$ ,
$\vec{a} + \vec{b} = 2 \vec{m}$
$\vec{n} = \frac{\vec{c} + \vec{d}}{2}$
$\vec{c} + \vec{d} = 2 \vec{n}$
$\vec{AD} = (\vec{d} - \vec{a}), \vec{BC} = (\vec{c} - \vec{b})$
Given,
$\vec{AD} + \vec{BC} = t (\overset{―}{MN})$
$(\vec{d} - \vec{a}) + (\vec{c} - \vec{b}) = t (\vec{n} - \vec{m})$
$\vec{d} - \vec{a} + \vec{c} - \vec{b} = t (\vec{n} - \vec{m})$
$(\vec{d} + \vec{c}) - (\vec{a} + \vec{b}) = t (\vec{n} - \vec{m})$
$2 \vec{n} - 2 \vec{m} = t (\vec{n} - \vec{m})$
$2 (\vec{n} - \vec{m}) = t (\vec{n} - \vec{m})$
So, $t = 2$

MHT CET-2020