QBManager

Vector Algebra

87671 If $P (1, 2, 3), R (4, 5, - 1)$ are the vertices and $G (2, 3, - 1)$ is the centroid of $△ P Q R$ , then coordinates of midpoint of $P Q$ are

1

(1, 2, 1)

2

(1, 2, 2)

3

(1, - 2, - 1)

4

(1, 2, - 1)

Explanation:

(D): Let,
$\vec{p}, \vec{q}, \vec{r}$ and $\vec{g}$ be the position vectors of $P, Q, R$ and $G$
respectively.
Given,
$\vec{p} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\vec{r} = 4 \hat{i} + 5 \hat{j} - \hat{k}$
$\vec{g} = 2 \hat{i} + 3 \hat{j} - \hat{k}$
We know that, Centroid,
$\vec{g} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$
$3 (2 \hat{i} + 3 \hat{j} - \hat{k}) = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \vec{q} + (4 \hat{i} + 5 \hat{j} - \hat{k})$
$6 \hat{i} + 9 \hat{j} - 3 \hat{k} = 5 \hat{i} + 7 \hat{j} + 2 \hat{k} + \vec{q}$
$\vec{q} = \hat{i} + 2 \hat{j} - 5 \hat{k}$
Co-ordinates of mid-point $P$ and $Q$ is,
$= (\frac{1 + 1}{2}, \frac{2 + 2}{2}, \frac{3 - 5}{2}) = (1, 2, - 1)$
Hence, the mid-point of PQ $(1, 2, - 1)$ .

MHT CET-2019

Vector Algebra

87672 If $A, B, C$ and $D$ are $(3, 7, 4), (5, - 2, 3), (- 4, 5, 6)$ and $(1, 2, 3)$ respectively, then the volume of the parallelopiped with $A B, A C$ and $A D$ as the coterminus edges, is (in cubic units)

1 92

2 94

3 91

4 93

Explanation:

(A) : Given,
$A = (3, 7, 4), B = (5, - 2, 3), C = (- 4, 5, 6), D = (1, 2, 3)$
Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be the position vectors of $A, B, C, D$ respectively.
$\vec{a} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{b} = 5 \hat{i} - 2 \hat{j} + 3 \hat{k}$
$\vec{c} = - 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$
$\vec{d} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$AB = \vec{b} - \vec{a}$
$\vec{AB} = (5 \hat{i} - 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AB} = 2 \vec{i} - 9 \hat{j} - \hat{k}$
$\vec{AD} = \vec{d} - \vec{a}$
$\vec{AD} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AD} = - 2 \hat{i} - 5 \hat{j} - \hat{k}$
$\vec{AC} = \vec{c} - \vec{a}$
$\vec{AC} = (- 4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AC} = - 7 \hat{i} - 2 \hat{j} + 2 \hat{k}$
Volume of parallelepiped,
$| \vec{AB} \vec{AC} \vec{AD} | = | \begin{array}{ccc} 2 & - 9 & - 1 \\ - 7 & - 2 & 2 \\ - 2 & - 5 & - 1 \end{array} |$
$= | 2 (2 + 10) + 9 (7 + 4) - 1 (35 - 4) | = | 24 + 99 - 31 | = | 92 |$
$= 92 cubic units$

MHT CET-2019

Vector Algebra

87673 If $G (3, - 5, r)$ is centroid of triangle $A B C$ where $A (7, - 8, 1), B (p, q, 5)$ and $C (q + 1, 5 p, 0)$ are vertices of a triangle then values of $p, q, r$ are respectively

1

- 4, 5, 4

2

6, 5, 4

3

- 3, 4, 3

4

- 2, 3, 2

Explanation:

(D) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{g}$ be the position vectors of $A, B, C$ and $G$ respectively.
$\vec{a} = 7 \hat{i} - 8 \hat{j} + \hat{k},$
$\vec{b} = p \hat{i} + q \hat{j} + 5 \hat{k}$
And, $\vec{c} = (q + 1) \hat{i} + 5 \hat{p} \hat{j} + 0 \hat{k}$
$\vec{g} = 3 \hat{i} - 5 \hat{j} + r \hat{k}$
By centroid formula,
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$3 \vec{g} = \vec{a} + \vec{b} + \vec{c}$
$3 (3 \hat{i} - 5 \hat{j} + r) = (7 \hat{i} - 8 \hat{j} + \hat{k}) + (p \hat{i} + q \hat{j} + 5 \hat{k})$
$+ [(q + 1) \hat{i} + 5 p \hat{j}]$
$9 \hat{i} - 15 \hat{j} + 3 r \hat{k} = (8 + p + q) \hat{i} + (- 8 + q + 5 p) \hat{j} + (6) \hat{k}$
By equality of vectors,
$8 + p + q = 9, - 8 + q + 5 p = - 15$ and $3 r = 6$
$∴ q = 1 - p \Rightarrow - 8 + 1 - p + 5 p = - 15 \Rightarrow 4 p - 7 = - 15 \Rightarrow p = - 2$
$∴ q = 1 - (- 2) = 3$ and $6 = 3 r \Rightarrow r = 2$
So, $(p, q, r) = (- 2, 3, 2)$

MHT CET-2019

Vector Algebra

87674 If $G (\vec{g}), H (\vec{h})$ and $P (\vec{p})$ are centroid, orthocentere and circumcentere of a triangle and $\vec{x} + y \vec{h} + z g = 0$ , then $(x, y, z) =$

1

1, 1, - 2

2

2, 1, - 3

3

1, 3, - 4

4

2, 3, - 5

Explanation:

(B) :
original image
Centroid (G) $= \vec{g}$
Orthocenter $(H) = \vec{h}$
Circumcenter $(P) = \vec{p}$
We know that,
Section formula,
$\vec{g} = \frac{2 \vec{p} + 1 \cdot \vec{h}}{2 + 1}$
$3 \vec{g} = 2 \vec{p} + \vec{h}$
$2 \vec{p} + \vec{h} - 3 \vec{g} = 0$
$x p + yh + z \vec{g} = 0$
On comparing equation (i) and (ii), we get-
$x = 2, y = 1, z = - 3$
So, $(x, y, z) = (2, 1, - 3)$

MHT CET-2016

Vector Algebra

87675 If $P$ is orthocenter, $Q$ is circumcentre and $G$ is centroid of $△ ABC$ , then $\overset{―}{QP} =$

1

3 \vec{QG}

2

2 \vec{QG}

3

\overset{―}{QG}

4

4 \vec{QG}

Explanation:

(A) : Let,
$\vec{p}$ and $\vec{g}$ be the position vectors of $P$ and $G$ with respect to the circumcentre $Q$ .
$\vec{QP} = \vec{p}$ and $\vec{QG} = \vec{g}$
We know that,
$G$ divides segment $\overset{―}{QP}$ internally ratio of 1:2
$\frac{ma + n \vec{b}}{m + n} = \vec{g}$
$∴ \vec{g} = \frac{1 \cdot \vec{p} + 2 \vec{q}}{1 + 2} = \frac{\vec{p}}{3} (∵ q = 0 where q is reference$
$point) \vec{p} = 3 \vec{g}$
$∴ \vec{QP} = 3 \vec{QG}$

MHT CET-2012

Vector Algebra

87671 If $P (1, 2, 3), R (4, 5, - 1)$ are the vertices and $G (2, 3, - 1)$ is the centroid of $△ P Q R$ , then coordinates of midpoint of $P Q$ are

1

(1, 2, 1)

2

(1, 2, 2)

3

(1, - 2, - 1)

4

(1, 2, - 1)

Explanation:

(D): Let,
$\vec{p}, \vec{q}, \vec{r}$ and $\vec{g}$ be the position vectors of $P, Q, R$ and $G$
respectively.
Given,
$\vec{p} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\vec{r} = 4 \hat{i} + 5 \hat{j} - \hat{k}$
$\vec{g} = 2 \hat{i} + 3 \hat{j} - \hat{k}$
We know that, Centroid,
$\vec{g} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$
$3 (2 \hat{i} + 3 \hat{j} - \hat{k}) = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \vec{q} + (4 \hat{i} + 5 \hat{j} - \hat{k})$
$6 \hat{i} + 9 \hat{j} - 3 \hat{k} = 5 \hat{i} + 7 \hat{j} + 2 \hat{k} + \vec{q}$
$\vec{q} = \hat{i} + 2 \hat{j} - 5 \hat{k}$
Co-ordinates of mid-point $P$ and $Q$ is,
$= (\frac{1 + 1}{2}, \frac{2 + 2}{2}, \frac{3 - 5}{2}) = (1, 2, - 1)$
Hence, the mid-point of PQ $(1, 2, - 1)$ .

MHT CET-2019

Vector Algebra

87672 If $A, B, C$ and $D$ are $(3, 7, 4), (5, - 2, 3), (- 4, 5, 6)$ and $(1, 2, 3)$ respectively, then the volume of the parallelopiped with $A B, A C$ and $A D$ as the coterminus edges, is (in cubic units)

1 92

2 94

3 91

4 93

Explanation:

(A) : Given,
$A = (3, 7, 4), B = (5, - 2, 3), C = (- 4, 5, 6), D = (1, 2, 3)$
Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be the position vectors of $A, B, C, D$ respectively.
$\vec{a} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{b} = 5 \hat{i} - 2 \hat{j} + 3 \hat{k}$
$\vec{c} = - 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$
$\vec{d} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$AB = \vec{b} - \vec{a}$
$\vec{AB} = (5 \hat{i} - 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AB} = 2 \vec{i} - 9 \hat{j} - \hat{k}$
$\vec{AD} = \vec{d} - \vec{a}$
$\vec{AD} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AD} = - 2 \hat{i} - 5 \hat{j} - \hat{k}$
$\vec{AC} = \vec{c} - \vec{a}$
$\vec{AC} = (- 4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AC} = - 7 \hat{i} - 2 \hat{j} + 2 \hat{k}$
Volume of parallelepiped,
$| \vec{AB} \vec{AC} \vec{AD} | = | \begin{array}{ccc} 2 & - 9 & - 1 \\ - 7 & - 2 & 2 \\ - 2 & - 5 & - 1 \end{array} |$
$= | 2 (2 + 10) + 9 (7 + 4) - 1 (35 - 4) | = | 24 + 99 - 31 | = | 92 |$
$= 92 cubic units$

MHT CET-2019

Vector Algebra

87673 If $G (3, - 5, r)$ is centroid of triangle $A B C$ where $A (7, - 8, 1), B (p, q, 5)$ and $C (q + 1, 5 p, 0)$ are vertices of a triangle then values of $p, q, r$ are respectively

1

- 4, 5, 4

2

6, 5, 4

3

- 3, 4, 3

4

- 2, 3, 2

Explanation:

(D) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{g}$ be the position vectors of $A, B, C$ and $G$ respectively.
$\vec{a} = 7 \hat{i} - 8 \hat{j} + \hat{k},$
$\vec{b} = p \hat{i} + q \hat{j} + 5 \hat{k}$
And, $\vec{c} = (q + 1) \hat{i} + 5 \hat{p} \hat{j} + 0 \hat{k}$
$\vec{g} = 3 \hat{i} - 5 \hat{j} + r \hat{k}$
By centroid formula,
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$3 \vec{g} = \vec{a} + \vec{b} + \vec{c}$
$3 (3 \hat{i} - 5 \hat{j} + r) = (7 \hat{i} - 8 \hat{j} + \hat{k}) + (p \hat{i} + q \hat{j} + 5 \hat{k})$
$+ [(q + 1) \hat{i} + 5 p \hat{j}]$
$9 \hat{i} - 15 \hat{j} + 3 r \hat{k} = (8 + p + q) \hat{i} + (- 8 + q + 5 p) \hat{j} + (6) \hat{k}$
By equality of vectors,
$8 + p + q = 9, - 8 + q + 5 p = - 15$ and $3 r = 6$
$∴ q = 1 - p \Rightarrow - 8 + 1 - p + 5 p = - 15 \Rightarrow 4 p - 7 = - 15 \Rightarrow p = - 2$
$∴ q = 1 - (- 2) = 3$ and $6 = 3 r \Rightarrow r = 2$
So, $(p, q, r) = (- 2, 3, 2)$

MHT CET-2019

Vector Algebra

87674 If $G (\vec{g}), H (\vec{h})$ and $P (\vec{p})$ are centroid, orthocentere and circumcentere of a triangle and $\vec{x} + y \vec{h} + z g = 0$ , then $(x, y, z) =$

1

1, 1, - 2

2

2, 1, - 3

3

1, 3, - 4

4

2, 3, - 5

Explanation:

(B) :
original image
Centroid (G) $= \vec{g}$
Orthocenter $(H) = \vec{h}$
Circumcenter $(P) = \vec{p}$
We know that,
Section formula,
$\vec{g} = \frac{2 \vec{p} + 1 \cdot \vec{h}}{2 + 1}$
$3 \vec{g} = 2 \vec{p} + \vec{h}$
$2 \vec{p} + \vec{h} - 3 \vec{g} = 0$
$x p + yh + z \vec{g} = 0$
On comparing equation (i) and (ii), we get-
$x = 2, y = 1, z = - 3$
So, $(x, y, z) = (2, 1, - 3)$

MHT CET-2016

Vector Algebra

87675 If $P$ is orthocenter, $Q$ is circumcentre and $G$ is centroid of $△ ABC$ , then $\overset{―}{QP} =$

1

3 \vec{QG}

2

2 \vec{QG}

3

\overset{―}{QG}

4

4 \vec{QG}

Explanation:

(A) : Let,
$\vec{p}$ and $\vec{g}$ be the position vectors of $P$ and $G$ with respect to the circumcentre $Q$ .
$\vec{QP} = \vec{p}$ and $\vec{QG} = \vec{g}$
We know that,
$G$ divides segment $\overset{―}{QP}$ internally ratio of 1:2
$\frac{ma + n \vec{b}}{m + n} = \vec{g}$
$∴ \vec{g} = \frac{1 \cdot \vec{p} + 2 \vec{q}}{1 + 2} = \frac{\vec{p}}{3} (∵ q = 0 where q is reference$
$point) \vec{p} = 3 \vec{g}$
$∴ \vec{QP} = 3 \vec{QG}$

MHT CET-2012

Vector Algebra

87671 If $P (1, 2, 3), R (4, 5, - 1)$ are the vertices and $G (2, 3, - 1)$ is the centroid of $△ P Q R$ , then coordinates of midpoint of $P Q$ are

1

(1, 2, 1)

2

(1, 2, 2)

3

(1, - 2, - 1)

4

(1, 2, - 1)

Explanation:

(D): Let,
$\vec{p}, \vec{q}, \vec{r}$ and $\vec{g}$ be the position vectors of $P, Q, R$ and $G$
respectively.
Given,
$\vec{p} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\vec{r} = 4 \hat{i} + 5 \hat{j} - \hat{k}$
$\vec{g} = 2 \hat{i} + 3 \hat{j} - \hat{k}$
We know that, Centroid,
$\vec{g} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$
$3 (2 \hat{i} + 3 \hat{j} - \hat{k}) = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \vec{q} + (4 \hat{i} + 5 \hat{j} - \hat{k})$
$6 \hat{i} + 9 \hat{j} - 3 \hat{k} = 5 \hat{i} + 7 \hat{j} + 2 \hat{k} + \vec{q}$
$\vec{q} = \hat{i} + 2 \hat{j} - 5 \hat{k}$
Co-ordinates of mid-point $P$ and $Q$ is,
$= (\frac{1 + 1}{2}, \frac{2 + 2}{2}, \frac{3 - 5}{2}) = (1, 2, - 1)$
Hence, the mid-point of PQ $(1, 2, - 1)$ .

MHT CET-2019

Vector Algebra

87672 If $A, B, C$ and $D$ are $(3, 7, 4), (5, - 2, 3), (- 4, 5, 6)$ and $(1, 2, 3)$ respectively, then the volume of the parallelopiped with $A B, A C$ and $A D$ as the coterminus edges, is (in cubic units)

1 92

2 94

3 91

4 93

Explanation:

(A) : Given,
$A = (3, 7, 4), B = (5, - 2, 3), C = (- 4, 5, 6), D = (1, 2, 3)$
Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be the position vectors of $A, B, C, D$ respectively.
$\vec{a} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{b} = 5 \hat{i} - 2 \hat{j} + 3 \hat{k}$
$\vec{c} = - 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$
$\vec{d} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$AB = \vec{b} - \vec{a}$
$\vec{AB} = (5 \hat{i} - 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AB} = 2 \vec{i} - 9 \hat{j} - \hat{k}$
$\vec{AD} = \vec{d} - \vec{a}$
$\vec{AD} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AD} = - 2 \hat{i} - 5 \hat{j} - \hat{k}$
$\vec{AC} = \vec{c} - \vec{a}$
$\vec{AC} = (- 4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AC} = - 7 \hat{i} - 2 \hat{j} + 2 \hat{k}$
Volume of parallelepiped,
$| \vec{AB} \vec{AC} \vec{AD} | = | \begin{array}{ccc} 2 & - 9 & - 1 \\ - 7 & - 2 & 2 \\ - 2 & - 5 & - 1 \end{array} |$
$= | 2 (2 + 10) + 9 (7 + 4) - 1 (35 - 4) | = | 24 + 99 - 31 | = | 92 |$
$= 92 cubic units$

MHT CET-2019

Vector Algebra

87673 If $G (3, - 5, r)$ is centroid of triangle $A B C$ where $A (7, - 8, 1), B (p, q, 5)$ and $C (q + 1, 5 p, 0)$ are vertices of a triangle then values of $p, q, r$ are respectively

1

- 4, 5, 4

2

6, 5, 4

3

- 3, 4, 3

4

- 2, 3, 2

Explanation:

(D) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{g}$ be the position vectors of $A, B, C$ and $G$ respectively.
$\vec{a} = 7 \hat{i} - 8 \hat{j} + \hat{k},$
$\vec{b} = p \hat{i} + q \hat{j} + 5 \hat{k}$
And, $\vec{c} = (q + 1) \hat{i} + 5 \hat{p} \hat{j} + 0 \hat{k}$
$\vec{g} = 3 \hat{i} - 5 \hat{j} + r \hat{k}$
By centroid formula,
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$3 \vec{g} = \vec{a} + \vec{b} + \vec{c}$
$3 (3 \hat{i} - 5 \hat{j} + r) = (7 \hat{i} - 8 \hat{j} + \hat{k}) + (p \hat{i} + q \hat{j} + 5 \hat{k})$
$+ [(q + 1) \hat{i} + 5 p \hat{j}]$
$9 \hat{i} - 15 \hat{j} + 3 r \hat{k} = (8 + p + q) \hat{i} + (- 8 + q + 5 p) \hat{j} + (6) \hat{k}$
By equality of vectors,
$8 + p + q = 9, - 8 + q + 5 p = - 15$ and $3 r = 6$
$∴ q = 1 - p \Rightarrow - 8 + 1 - p + 5 p = - 15 \Rightarrow 4 p - 7 = - 15 \Rightarrow p = - 2$
$∴ q = 1 - (- 2) = 3$ and $6 = 3 r \Rightarrow r = 2$
So, $(p, q, r) = (- 2, 3, 2)$

MHT CET-2019

Vector Algebra

87674 If $G (\vec{g}), H (\vec{h})$ and $P (\vec{p})$ are centroid, orthocentere and circumcentere of a triangle and $\vec{x} + y \vec{h} + z g = 0$ , then $(x, y, z) =$

1

1, 1, - 2

2

2, 1, - 3

3

1, 3, - 4

4

2, 3, - 5

Explanation:

(B) :
original image
Centroid (G) $= \vec{g}$
Orthocenter $(H) = \vec{h}$
Circumcenter $(P) = \vec{p}$
We know that,
Section formula,
$\vec{g} = \frac{2 \vec{p} + 1 \cdot \vec{h}}{2 + 1}$
$3 \vec{g} = 2 \vec{p} + \vec{h}$
$2 \vec{p} + \vec{h} - 3 \vec{g} = 0$
$x p + yh + z \vec{g} = 0$
On comparing equation (i) and (ii), we get-
$x = 2, y = 1, z = - 3$
So, $(x, y, z) = (2, 1, - 3)$

MHT CET-2016

Vector Algebra

87675 If $P$ is orthocenter, $Q$ is circumcentre and $G$ is centroid of $△ ABC$ , then $\overset{―}{QP} =$

1

3 \vec{QG}

2

2 \vec{QG}

3

\overset{―}{QG}

4

4 \vec{QG}

Explanation:

(A) : Let,
$\vec{p}$ and $\vec{g}$ be the position vectors of $P$ and $G$ with respect to the circumcentre $Q$ .
$\vec{QP} = \vec{p}$ and $\vec{QG} = \vec{g}$
We know that,
$G$ divides segment $\overset{―}{QP}$ internally ratio of 1:2
$\frac{ma + n \vec{b}}{m + n} = \vec{g}$
$∴ \vec{g} = \frac{1 \cdot \vec{p} + 2 \vec{q}}{1 + 2} = \frac{\vec{p}}{3} (∵ q = 0 where q is reference$
$point) \vec{p} = 3 \vec{g}$
$∴ \vec{QP} = 3 \vec{QG}$

MHT CET-2012

Vector Algebra

87671 If $P (1, 2, 3), R (4, 5, - 1)$ are the vertices and $G (2, 3, - 1)$ is the centroid of $△ P Q R$ , then coordinates of midpoint of $P Q$ are

1

(1, 2, 1)

2

(1, 2, 2)

3

(1, - 2, - 1)

4

(1, 2, - 1)

Explanation:

(D): Let,
$\vec{p}, \vec{q}, \vec{r}$ and $\vec{g}$ be the position vectors of $P, Q, R$ and $G$
respectively.
Given,
$\vec{p} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\vec{r} = 4 \hat{i} + 5 \hat{j} - \hat{k}$
$\vec{g} = 2 \hat{i} + 3 \hat{j} - \hat{k}$
We know that, Centroid,
$\vec{g} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$
$3 (2 \hat{i} + 3 \hat{j} - \hat{k}) = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \vec{q} + (4 \hat{i} + 5 \hat{j} - \hat{k})$
$6 \hat{i} + 9 \hat{j} - 3 \hat{k} = 5 \hat{i} + 7 \hat{j} + 2 \hat{k} + \vec{q}$
$\vec{q} = \hat{i} + 2 \hat{j} - 5 \hat{k}$
Co-ordinates of mid-point $P$ and $Q$ is,
$= (\frac{1 + 1}{2}, \frac{2 + 2}{2}, \frac{3 - 5}{2}) = (1, 2, - 1)$
Hence, the mid-point of PQ $(1, 2, - 1)$ .

MHT CET-2019

Vector Algebra

87672 If $A, B, C$ and $D$ are $(3, 7, 4), (5, - 2, 3), (- 4, 5, 6)$ and $(1, 2, 3)$ respectively, then the volume of the parallelopiped with $A B, A C$ and $A D$ as the coterminus edges, is (in cubic units)

1 92

2 94

3 91

4 93

Explanation:

(A) : Given,
$A = (3, 7, 4), B = (5, - 2, 3), C = (- 4, 5, 6), D = (1, 2, 3)$
Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be the position vectors of $A, B, C, D$ respectively.
$\vec{a} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{b} = 5 \hat{i} - 2 \hat{j} + 3 \hat{k}$
$\vec{c} = - 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$
$\vec{d} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$AB = \vec{b} - \vec{a}$
$\vec{AB} = (5 \hat{i} - 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AB} = 2 \vec{i} - 9 \hat{j} - \hat{k}$
$\vec{AD} = \vec{d} - \vec{a}$
$\vec{AD} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AD} = - 2 \hat{i} - 5 \hat{j} - \hat{k}$
$\vec{AC} = \vec{c} - \vec{a}$
$\vec{AC} = (- 4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AC} = - 7 \hat{i} - 2 \hat{j} + 2 \hat{k}$
Volume of parallelepiped,
$| \vec{AB} \vec{AC} \vec{AD} | = | \begin{array}{ccc} 2 & - 9 & - 1 \\ - 7 & - 2 & 2 \\ - 2 & - 5 & - 1 \end{array} |$
$= | 2 (2 + 10) + 9 (7 + 4) - 1 (35 - 4) | = | 24 + 99 - 31 | = | 92 |$
$= 92 cubic units$

MHT CET-2019

Vector Algebra

87673 If $G (3, - 5, r)$ is centroid of triangle $A B C$ where $A (7, - 8, 1), B (p, q, 5)$ and $C (q + 1, 5 p, 0)$ are vertices of a triangle then values of $p, q, r$ are respectively

1

- 4, 5, 4

2

6, 5, 4

3

- 3, 4, 3

4

- 2, 3, 2

Explanation:

(D) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{g}$ be the position vectors of $A, B, C$ and $G$ respectively.
$\vec{a} = 7 \hat{i} - 8 \hat{j} + \hat{k},$
$\vec{b} = p \hat{i} + q \hat{j} + 5 \hat{k}$
And, $\vec{c} = (q + 1) \hat{i} + 5 \hat{p} \hat{j} + 0 \hat{k}$
$\vec{g} = 3 \hat{i} - 5 \hat{j} + r \hat{k}$
By centroid formula,
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$3 \vec{g} = \vec{a} + \vec{b} + \vec{c}$
$3 (3 \hat{i} - 5 \hat{j} + r) = (7 \hat{i} - 8 \hat{j} + \hat{k}) + (p \hat{i} + q \hat{j} + 5 \hat{k})$
$+ [(q + 1) \hat{i} + 5 p \hat{j}]$
$9 \hat{i} - 15 \hat{j} + 3 r \hat{k} = (8 + p + q) \hat{i} + (- 8 + q + 5 p) \hat{j} + (6) \hat{k}$
By equality of vectors,
$8 + p + q = 9, - 8 + q + 5 p = - 15$ and $3 r = 6$
$∴ q = 1 - p \Rightarrow - 8 + 1 - p + 5 p = - 15 \Rightarrow 4 p - 7 = - 15 \Rightarrow p = - 2$
$∴ q = 1 - (- 2) = 3$ and $6 = 3 r \Rightarrow r = 2$
So, $(p, q, r) = (- 2, 3, 2)$

MHT CET-2019

Vector Algebra

87674 If $G (\vec{g}), H (\vec{h})$ and $P (\vec{p})$ are centroid, orthocentere and circumcentere of a triangle and $\vec{x} + y \vec{h} + z g = 0$ , then $(x, y, z) =$

1

1, 1, - 2

2

2, 1, - 3

3

1, 3, - 4

4

2, 3, - 5

Explanation:

(B) :
original image
Centroid (G) $= \vec{g}$
Orthocenter $(H) = \vec{h}$
Circumcenter $(P) = \vec{p}$
We know that,
Section formula,
$\vec{g} = \frac{2 \vec{p} + 1 \cdot \vec{h}}{2 + 1}$
$3 \vec{g} = 2 \vec{p} + \vec{h}$
$2 \vec{p} + \vec{h} - 3 \vec{g} = 0$
$x p + yh + z \vec{g} = 0$
On comparing equation (i) and (ii), we get-
$x = 2, y = 1, z = - 3$
So, $(x, y, z) = (2, 1, - 3)$

MHT CET-2016

Vector Algebra

87675 If $P$ is orthocenter, $Q$ is circumcentre and $G$ is centroid of $△ ABC$ , then $\overset{―}{QP} =$

1

3 \vec{QG}

2

2 \vec{QG}

3

\overset{―}{QG}

4

4 \vec{QG}

Explanation:

(A) : Let,
$\vec{p}$ and $\vec{g}$ be the position vectors of $P$ and $G$ with respect to the circumcentre $Q$ .
$\vec{QP} = \vec{p}$ and $\vec{QG} = \vec{g}$
We know that,
$G$ divides segment $\overset{―}{QP}$ internally ratio of 1:2
$\frac{ma + n \vec{b}}{m + n} = \vec{g}$
$∴ \vec{g} = \frac{1 \cdot \vec{p} + 2 \vec{q}}{1 + 2} = \frac{\vec{p}}{3} (∵ q = 0 where q is reference$
$point) \vec{p} = 3 \vec{g}$
$∴ \vec{QP} = 3 \vec{QG}$

MHT CET-2012

Vector Algebra

87671 If $P (1, 2, 3), R (4, 5, - 1)$ are the vertices and $G (2, 3, - 1)$ is the centroid of $△ P Q R$ , then coordinates of midpoint of $P Q$ are

1

(1, 2, 1)

2

(1, 2, 2)

3

(1, - 2, - 1)

4

(1, 2, - 1)

Explanation:

(D): Let,
$\vec{p}, \vec{q}, \vec{r}$ and $\vec{g}$ be the position vectors of $P, Q, R$ and $G$
respectively.
Given,
$\vec{p} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\vec{r} = 4 \hat{i} + 5 \hat{j} - \hat{k}$
$\vec{g} = 2 \hat{i} + 3 \hat{j} - \hat{k}$
We know that, Centroid,
$\vec{g} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$
$3 (2 \hat{i} + 3 \hat{j} - \hat{k}) = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \vec{q} + (4 \hat{i} + 5 \hat{j} - \hat{k})$
$6 \hat{i} + 9 \hat{j} - 3 \hat{k} = 5 \hat{i} + 7 \hat{j} + 2 \hat{k} + \vec{q}$
$\vec{q} = \hat{i} + 2 \hat{j} - 5 \hat{k}$
Co-ordinates of mid-point $P$ and $Q$ is,
$= (\frac{1 + 1}{2}, \frac{2 + 2}{2}, \frac{3 - 5}{2}) = (1, 2, - 1)$
Hence, the mid-point of PQ $(1, 2, - 1)$ .

MHT CET-2019

Vector Algebra

87672 If $A, B, C$ and $D$ are $(3, 7, 4), (5, - 2, 3), (- 4, 5, 6)$ and $(1, 2, 3)$ respectively, then the volume of the parallelopiped with $A B, A C$ and $A D$ as the coterminus edges, is (in cubic units)

1 92

2 94

3 91

4 93

Explanation:

(A) : Given,
$A = (3, 7, 4), B = (5, - 2, 3), C = (- 4, 5, 6), D = (1, 2, 3)$
Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be the position vectors of $A, B, C, D$ respectively.
$\vec{a} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{b} = 5 \hat{i} - 2 \hat{j} + 3 \hat{k}$
$\vec{c} = - 4 \hat{i} + 5 \hat{j} + 6 \hat{k}$
$\vec{d} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$AB = \vec{b} - \vec{a}$
$\vec{AB} = (5 \hat{i} - 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AB} = 2 \vec{i} - 9 \hat{j} - \hat{k}$
$\vec{AD} = \vec{d} - \vec{a}$
$\vec{AD} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AD} = - 2 \hat{i} - 5 \hat{j} - \hat{k}$
$\vec{AC} = \vec{c} - \vec{a}$
$\vec{AC} = (- 4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (3 \hat{i} + 7 \hat{j} + 4 \hat{k})$
$\vec{AC} = - 7 \hat{i} - 2 \hat{j} + 2 \hat{k}$
Volume of parallelepiped,
$| \vec{AB} \vec{AC} \vec{AD} | = | \begin{array}{ccc} 2 & - 9 & - 1 \\ - 7 & - 2 & 2 \\ - 2 & - 5 & - 1 \end{array} |$
$= | 2 (2 + 10) + 9 (7 + 4) - 1 (35 - 4) | = | 24 + 99 - 31 | = | 92 |$
$= 92 cubic units$

MHT CET-2019

Vector Algebra

87673 If $G (3, - 5, r)$ is centroid of triangle $A B C$ where $A (7, - 8, 1), B (p, q, 5)$ and $C (q + 1, 5 p, 0)$ are vertices of a triangle then values of $p, q, r$ are respectively

1

- 4, 5, 4

2

6, 5, 4

3

- 3, 4, 3

4

- 2, 3, 2

Explanation:

(D) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{g}$ be the position vectors of $A, B, C$ and $G$ respectively.
$\vec{a} = 7 \hat{i} - 8 \hat{j} + \hat{k},$
$\vec{b} = p \hat{i} + q \hat{j} + 5 \hat{k}$
And, $\vec{c} = (q + 1) \hat{i} + 5 \hat{p} \hat{j} + 0 \hat{k}$
$\vec{g} = 3 \hat{i} - 5 \hat{j} + r \hat{k}$
By centroid formula,
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
$3 \vec{g} = \vec{a} + \vec{b} + \vec{c}$
$3 (3 \hat{i} - 5 \hat{j} + r) = (7 \hat{i} - 8 \hat{j} + \hat{k}) + (p \hat{i} + q \hat{j} + 5 \hat{k})$
$+ [(q + 1) \hat{i} + 5 p \hat{j}]$
$9 \hat{i} - 15 \hat{j} + 3 r \hat{k} = (8 + p + q) \hat{i} + (- 8 + q + 5 p) \hat{j} + (6) \hat{k}$
By equality of vectors,
$8 + p + q = 9, - 8 + q + 5 p = - 15$ and $3 r = 6$
$∴ q = 1 - p \Rightarrow - 8 + 1 - p + 5 p = - 15 \Rightarrow 4 p - 7 = - 15 \Rightarrow p = - 2$
$∴ q = 1 - (- 2) = 3$ and $6 = 3 r \Rightarrow r = 2$
So, $(p, q, r) = (- 2, 3, 2)$

MHT CET-2019

Vector Algebra

87674 If $G (\vec{g}), H (\vec{h})$ and $P (\vec{p})$ are centroid, orthocentere and circumcentere of a triangle and $\vec{x} + y \vec{h} + z g = 0$ , then $(x, y, z) =$

1

1, 1, - 2

2

2, 1, - 3

3

1, 3, - 4

4

2, 3, - 5

Explanation:

(B) :
original image
Centroid (G) $= \vec{g}$
Orthocenter $(H) = \vec{h}$
Circumcenter $(P) = \vec{p}$
We know that,
Section formula,
$\vec{g} = \frac{2 \vec{p} + 1 \cdot \vec{h}}{2 + 1}$
$3 \vec{g} = 2 \vec{p} + \vec{h}$
$2 \vec{p} + \vec{h} - 3 \vec{g} = 0$
$x p + yh + z \vec{g} = 0$
On comparing equation (i) and (ii), we get-
$x = 2, y = 1, z = - 3$
So, $(x, y, z) = (2, 1, - 3)$

MHT CET-2016

Vector Algebra

87675 If $P$ is orthocenter, $Q$ is circumcentre and $G$ is centroid of $△ ABC$ , then $\overset{―}{QP} =$

1

3 \vec{QG}

2

2 \vec{QG}

3

\overset{―}{QG}

4

4 \vec{QG}

Explanation:

(A) : Let,
$\vec{p}$ and $\vec{g}$ be the position vectors of $P$ and $G$ with respect to the circumcentre $Q$ .
$\vec{QP} = \vec{p}$ and $\vec{QG} = \vec{g}$
We know that,
$G$ divides segment $\overset{―}{QP}$ internally ratio of 1:2
$\frac{ma + n \vec{b}}{m + n} = \vec{g}$
$∴ \vec{g} = \frac{1 \cdot \vec{p} + 2 \vec{q}}{1 + 2} = \frac{\vec{p}}{3} (∵ q = 0 where q is reference$
$point) \vec{p} = 3 \vec{g}$
$∴ \vec{QP} = 3 \vec{QG}$

MHT CET-2012

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: