QBManager

Vector Algebra

87667 If $\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}, \vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$ represent the sides of triangle $A B C$ , then the length of median through $A$ is

1 6 units

2 5 units

3

\sqrt{6}

units

4

\sqrt{5}

units

Explanation:

(B)Given,
$\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$
$\vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$
Let $\vec{AD}$ is median Position vector of,
original image
$\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$
$\vec{AD} = \frac{(3 \hat{i} + 5 \hat{j} + 4 \hat{k}) + (5 \hat{i} - 5 \hat{j} + 2 \hat{k})}{2}$
$\vec{AD} = \frac{(3 + 5) \hat{i} + (5 - 5) \hat{j} + (4 + 2) \hat{k}}{2}$
$\vec{AD} = \frac{8 \hat{i} + 6 \hat{k}}{2}$
$\vec{AD} = 4 \hat{i} + 3 \hat{k}$
$∣ \vec{AD} = \sqrt{16 + 9} = \sqrt{25} = 5$
Hence, length of median is 5 units

MHT CET-2020

Vector Algebra

87668 If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are the position vector of the points
A, B, C, D respectively such that $3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = \vec{0}$ , then the position vector of the point of intersection of the line segments $A C$ and $B D$ is

1

\frac{\vec{a} + \vec{b}}{2}

2

\frac{3 \vec{a} + \vec{c}}{4}

3

\frac{\vec{b} + 3 \vec{d}}{4}

4

\frac{\vec{b} + 4 \vec{d}}{5}

Explanation:

(D) Given,
$3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = 0$
$Rearranging the term in equation,$
$3 \vec{a} + 2 \vec{c} = \vec{b} + 4 \vec{d}$
$∴ \frac{3 \vec{a} + 2 \vec{c}}{3 + 2} = \frac{\vec{b} + 4 d}{1 + 4} \Rightarrow So, \frac{3 \vec{a} + 2 \vec{c}}{5} = \frac{\vec{b} + 4 \vec{d}}{5}$

MHT CET-2020

Vector Algebra

87669 If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the points $A (1, 3, 0), B (2, 5, 0), C (4, 2, 0)$ respectively and $\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$ , then value of $t_{1} t_{2} =$

1 160

2 -16

3 16

4 -160

Explanation:

(D) : Given,
$\vec{c} = 4 \hat{i} + 2 \hat{j} + 0 \hat{k}$
$\vec{a} = \hat{i} + 3 \hat{j} + 0 \hat{k}$
$\vec{b} = 2 \hat{i} + 5 \hat{j} + 0 \hat{k}$
From given conditions,
$\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$
$(4 \hat{i} + 2 \hat{j} + 0 \hat{k}) = t_{1} (\hat{i} + 3 \hat{j} + 0 \hat{k}) + t_{2} (2 \hat{i} + 5 \hat{j} + 0 \hat{k})$
$∴ = (t_{1} + 2 t_{2}) \hat{i} + (3 t_{1} + 5 t_{2}) \hat{j}$
$∴ t_{1} + 2 t_{2} = 4 and 3 t_{1} + 5 t_{2} = 2$
Solving, we get $t_{2} = 10$ and $t_{1} = - 16$
So, $t_{1} t_{2} = - 160$

MHT CET-2020

Vector Algebra

87670 $A \equiv (2, 3, - 2)$ and $B \equiv (4, 1, - 2)$ are two vertices of $△ A B C . P, Q$ and $R$ are the midpoints of $A B$ , $B C$ and $A C$ respectively. The coordinates of $R$ are $(\frac{5}{2}, \frac{5}{2}, - \frac{7}{2})$ . Then the centroid of $△ P Q R$ is

1

(3, - 2, - 3)

2

(3, 2, - 3)

3

(3, 2, 3)

4

(\frac{3}{2}, - \frac{7}{2}, \frac{3}{2})

Explanation:

(B) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{p}, \vec{q}, \vec{r}$ be the position vectors of $A, B, C, P, Q$ , $R$ respectively.
$\vec{a} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k},$
$\vec{b} = 4 \hat{i} + \hat{j} - 2 \hat{k},$
$\vec{r} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k}$
Consider, $\vec{c} = xi + y \hat{j} + z \hat{k}$
Then, from question -
$\frac{\vec{a} + \vec{c}}{2} = \vec{r}$
$\frac{(2 \hat{i} + 3 \hat{j} - 2 \hat{k}) + (x \hat{i} + y \hat{j} + z \hat{k})}{2} = (\frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k})$
$(2 + x) \hat{i} + (3 + y) \hat{j} + (z - 2) \hat{k} = 5 \hat{i} + 5 \hat{j} - 7 \hat{k}$
On comparing both side, we get -
$x = 3, y = 2, z = - 5$
Then, $\vec{c} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$
As the centroid of the triangle formed by joining the mid points of the sides of a given triangle coincides with the centroid of the given triangle.
$∴$ Centroid
$of △ PQR = (\frac{2 + 4 + 3}{3}, \frac{3 + 1 + 2}{3}, \frac{- 2 - 2 - 5}{3}) = (3, 2, - 3)$

MHT CET-2019

Vector Algebra

87667 If $\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}, \vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$ represent the sides of triangle $A B C$ , then the length of median through $A$ is

1 6 units

2 5 units

3

\sqrt{6}

units

4

\sqrt{5}

units

Explanation:

(B)Given,
$\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$
$\vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$
Let $\vec{AD}$ is median Position vector of,
original image
$\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$
$\vec{AD} = \frac{(3 \hat{i} + 5 \hat{j} + 4 \hat{k}) + (5 \hat{i} - 5 \hat{j} + 2 \hat{k})}{2}$
$\vec{AD} = \frac{(3 + 5) \hat{i} + (5 - 5) \hat{j} + (4 + 2) \hat{k}}{2}$
$\vec{AD} = \frac{8 \hat{i} + 6 \hat{k}}{2}$
$\vec{AD} = 4 \hat{i} + 3 \hat{k}$
$∣ \vec{AD} = \sqrt{16 + 9} = \sqrt{25} = 5$
Hence, length of median is 5 units

MHT CET-2020

Vector Algebra

87668 If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are the position vector of the points
A, B, C, D respectively such that $3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = \vec{0}$ , then the position vector of the point of intersection of the line segments $A C$ and $B D$ is

1

\frac{\vec{a} + \vec{b}}{2}

2

\frac{3 \vec{a} + \vec{c}}{4}

3

\frac{\vec{b} + 3 \vec{d}}{4}

4

\frac{\vec{b} + 4 \vec{d}}{5}

Explanation:

(D) Given,
$3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = 0$
$Rearranging the term in equation,$
$3 \vec{a} + 2 \vec{c} = \vec{b} + 4 \vec{d}$
$∴ \frac{3 \vec{a} + 2 \vec{c}}{3 + 2} = \frac{\vec{b} + 4 d}{1 + 4} \Rightarrow So, \frac{3 \vec{a} + 2 \vec{c}}{5} = \frac{\vec{b} + 4 \vec{d}}{5}$

MHT CET-2020

Vector Algebra

87669 If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the points $A (1, 3, 0), B (2, 5, 0), C (4, 2, 0)$ respectively and $\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$ , then value of $t_{1} t_{2} =$

1 160

2 -16

3 16

4 -160

Explanation:

(D) : Given,
$\vec{c} = 4 \hat{i} + 2 \hat{j} + 0 \hat{k}$
$\vec{a} = \hat{i} + 3 \hat{j} + 0 \hat{k}$
$\vec{b} = 2 \hat{i} + 5 \hat{j} + 0 \hat{k}$
From given conditions,
$\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$
$(4 \hat{i} + 2 \hat{j} + 0 \hat{k}) = t_{1} (\hat{i} + 3 \hat{j} + 0 \hat{k}) + t_{2} (2 \hat{i} + 5 \hat{j} + 0 \hat{k})$
$∴ = (t_{1} + 2 t_{2}) \hat{i} + (3 t_{1} + 5 t_{2}) \hat{j}$
$∴ t_{1} + 2 t_{2} = 4 and 3 t_{1} + 5 t_{2} = 2$
Solving, we get $t_{2} = 10$ and $t_{1} = - 16$
So, $t_{1} t_{2} = - 160$

MHT CET-2020

Vector Algebra

87670 $A \equiv (2, 3, - 2)$ and $B \equiv (4, 1, - 2)$ are two vertices of $△ A B C . P, Q$ and $R$ are the midpoints of $A B$ , $B C$ and $A C$ respectively. The coordinates of $R$ are $(\frac{5}{2}, \frac{5}{2}, - \frac{7}{2})$ . Then the centroid of $△ P Q R$ is

1

(3, - 2, - 3)

2

(3, 2, - 3)

3

(3, 2, 3)

4

(\frac{3}{2}, - \frac{7}{2}, \frac{3}{2})

Explanation:

(B) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{p}, \vec{q}, \vec{r}$ be the position vectors of $A, B, C, P, Q$ , $R$ respectively.
$\vec{a} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k},$
$\vec{b} = 4 \hat{i} + \hat{j} - 2 \hat{k},$
$\vec{r} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k}$
Consider, $\vec{c} = xi + y \hat{j} + z \hat{k}$
Then, from question -
$\frac{\vec{a} + \vec{c}}{2} = \vec{r}$
$\frac{(2 \hat{i} + 3 \hat{j} - 2 \hat{k}) + (x \hat{i} + y \hat{j} + z \hat{k})}{2} = (\frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k})$
$(2 + x) \hat{i} + (3 + y) \hat{j} + (z - 2) \hat{k} = 5 \hat{i} + 5 \hat{j} - 7 \hat{k}$
On comparing both side, we get -
$x = 3, y = 2, z = - 5$
Then, $\vec{c} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$
As the centroid of the triangle formed by joining the mid points of the sides of a given triangle coincides with the centroid of the given triangle.
$∴$ Centroid
$of △ PQR = (\frac{2 + 4 + 3}{3}, \frac{3 + 1 + 2}{3}, \frac{- 2 - 2 - 5}{3}) = (3, 2, - 3)$

MHT CET-2019

Vector Algebra

87667 If $\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}, \vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$ represent the sides of triangle $A B C$ , then the length of median through $A$ is

1 6 units

2 5 units

3

\sqrt{6}

units

4

\sqrt{5}

units

Explanation:

(B)Given,
$\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$
$\vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$
Let $\vec{AD}$ is median Position vector of,
original image
$\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$
$\vec{AD} = \frac{(3 \hat{i} + 5 \hat{j} + 4 \hat{k}) + (5 \hat{i} - 5 \hat{j} + 2 \hat{k})}{2}$
$\vec{AD} = \frac{(3 + 5) \hat{i} + (5 - 5) \hat{j} + (4 + 2) \hat{k}}{2}$
$\vec{AD} = \frac{8 \hat{i} + 6 \hat{k}}{2}$
$\vec{AD} = 4 \hat{i} + 3 \hat{k}$
$∣ \vec{AD} = \sqrt{16 + 9} = \sqrt{25} = 5$
Hence, length of median is 5 units

MHT CET-2020

Vector Algebra

87668 If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are the position vector of the points
A, B, C, D respectively such that $3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = \vec{0}$ , then the position vector of the point of intersection of the line segments $A C$ and $B D$ is

1

\frac{\vec{a} + \vec{b}}{2}

2

\frac{3 \vec{a} + \vec{c}}{4}

3

\frac{\vec{b} + 3 \vec{d}}{4}

4

\frac{\vec{b} + 4 \vec{d}}{5}

Explanation:

(D) Given,
$3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = 0$
$Rearranging the term in equation,$
$3 \vec{a} + 2 \vec{c} = \vec{b} + 4 \vec{d}$
$∴ \frac{3 \vec{a} + 2 \vec{c}}{3 + 2} = \frac{\vec{b} + 4 d}{1 + 4} \Rightarrow So, \frac{3 \vec{a} + 2 \vec{c}}{5} = \frac{\vec{b} + 4 \vec{d}}{5}$

MHT CET-2020

Vector Algebra

87669 If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the points $A (1, 3, 0), B (2, 5, 0), C (4, 2, 0)$ respectively and $\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$ , then value of $t_{1} t_{2} =$

1 160

2 -16

3 16

4 -160

Explanation:

(D) : Given,
$\vec{c} = 4 \hat{i} + 2 \hat{j} + 0 \hat{k}$
$\vec{a} = \hat{i} + 3 \hat{j} + 0 \hat{k}$
$\vec{b} = 2 \hat{i} + 5 \hat{j} + 0 \hat{k}$
From given conditions,
$\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$
$(4 \hat{i} + 2 \hat{j} + 0 \hat{k}) = t_{1} (\hat{i} + 3 \hat{j} + 0 \hat{k}) + t_{2} (2 \hat{i} + 5 \hat{j} + 0 \hat{k})$
$∴ = (t_{1} + 2 t_{2}) \hat{i} + (3 t_{1} + 5 t_{2}) \hat{j}$
$∴ t_{1} + 2 t_{2} = 4 and 3 t_{1} + 5 t_{2} = 2$
Solving, we get $t_{2} = 10$ and $t_{1} = - 16$
So, $t_{1} t_{2} = - 160$

MHT CET-2020

Vector Algebra

87670 $A \equiv (2, 3, - 2)$ and $B \equiv (4, 1, - 2)$ are two vertices of $△ A B C . P, Q$ and $R$ are the midpoints of $A B$ , $B C$ and $A C$ respectively. The coordinates of $R$ are $(\frac{5}{2}, \frac{5}{2}, - \frac{7}{2})$ . Then the centroid of $△ P Q R$ is

1

(3, - 2, - 3)

2

(3, 2, - 3)

3

(3, 2, 3)

4

(\frac{3}{2}, - \frac{7}{2}, \frac{3}{2})

Explanation:

(B) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{p}, \vec{q}, \vec{r}$ be the position vectors of $A, B, C, P, Q$ , $R$ respectively.
$\vec{a} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k},$
$\vec{b} = 4 \hat{i} + \hat{j} - 2 \hat{k},$
$\vec{r} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k}$
Consider, $\vec{c} = xi + y \hat{j} + z \hat{k}$
Then, from question -
$\frac{\vec{a} + \vec{c}}{2} = \vec{r}$
$\frac{(2 \hat{i} + 3 \hat{j} - 2 \hat{k}) + (x \hat{i} + y \hat{j} + z \hat{k})}{2} = (\frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k})$
$(2 + x) \hat{i} + (3 + y) \hat{j} + (z - 2) \hat{k} = 5 \hat{i} + 5 \hat{j} - 7 \hat{k}$
On comparing both side, we get -
$x = 3, y = 2, z = - 5$
Then, $\vec{c} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$
As the centroid of the triangle formed by joining the mid points of the sides of a given triangle coincides with the centroid of the given triangle.
$∴$ Centroid
$of △ PQR = (\frac{2 + 4 + 3}{3}, \frac{3 + 1 + 2}{3}, \frac{- 2 - 2 - 5}{3}) = (3, 2, - 3)$

MHT CET-2019

Vector Algebra

87667 If $\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}, \vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$ represent the sides of triangle $A B C$ , then the length of median through $A$ is

1 6 units

2 5 units

3

\sqrt{6}

units

4

\sqrt{5}

units

Explanation:

(B)Given,
$\vec{AB} = 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$
$\vec{AC} = 5 \hat{i} - 5 \hat{j} + 2 \hat{k}$
Let $\vec{AD}$ is median Position vector of,
original image
$\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$
$\vec{AD} = \frac{(3 \hat{i} + 5 \hat{j} + 4 \hat{k}) + (5 \hat{i} - 5 \hat{j} + 2 \hat{k})}{2}$
$\vec{AD} = \frac{(3 + 5) \hat{i} + (5 - 5) \hat{j} + (4 + 2) \hat{k}}{2}$
$\vec{AD} = \frac{8 \hat{i} + 6 \hat{k}}{2}$
$\vec{AD} = 4 \hat{i} + 3 \hat{k}$
$∣ \vec{AD} = \sqrt{16 + 9} = \sqrt{25} = 5$
Hence, length of median is 5 units

MHT CET-2020

Vector Algebra

87668 If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are the position vector of the points
A, B, C, D respectively such that $3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = \vec{0}$ , then the position vector of the point of intersection of the line segments $A C$ and $B D$ is

1

\frac{\vec{a} + \vec{b}}{2}

2

\frac{3 \vec{a} + \vec{c}}{4}

3

\frac{\vec{b} + 3 \vec{d}}{4}

4

\frac{\vec{b} + 4 \vec{d}}{5}

Explanation:

(D) Given,
$3 \vec{a} - \vec{b} + 2 \vec{c} - 4 \vec{d} = 0$
$Rearranging the term in equation,$
$3 \vec{a} + 2 \vec{c} = \vec{b} + 4 \vec{d}$
$∴ \frac{3 \vec{a} + 2 \vec{c}}{3 + 2} = \frac{\vec{b} + 4 d}{1 + 4} \Rightarrow So, \frac{3 \vec{a} + 2 \vec{c}}{5} = \frac{\vec{b} + 4 \vec{d}}{5}$

MHT CET-2020

Vector Algebra

87669 If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the points $A (1, 3, 0), B (2, 5, 0), C (4, 2, 0)$ respectively and $\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$ , then value of $t_{1} t_{2} =$

1 160

2 -16

3 16

4 -160

Explanation:

(D) : Given,
$\vec{c} = 4 \hat{i} + 2 \hat{j} + 0 \hat{k}$
$\vec{a} = \hat{i} + 3 \hat{j} + 0 \hat{k}$
$\vec{b} = 2 \hat{i} + 5 \hat{j} + 0 \hat{k}$
From given conditions,
$\vec{c} = t_{1} \vec{a} + t_{2} \vec{b}$
$(4 \hat{i} + 2 \hat{j} + 0 \hat{k}) = t_{1} (\hat{i} + 3 \hat{j} + 0 \hat{k}) + t_{2} (2 \hat{i} + 5 \hat{j} + 0 \hat{k})$
$∴ = (t_{1} + 2 t_{2}) \hat{i} + (3 t_{1} + 5 t_{2}) \hat{j}$
$∴ t_{1} + 2 t_{2} = 4 and 3 t_{1} + 5 t_{2} = 2$
Solving, we get $t_{2} = 10$ and $t_{1} = - 16$
So, $t_{1} t_{2} = - 160$

MHT CET-2020

Vector Algebra

87670 $A \equiv (2, 3, - 2)$ and $B \equiv (4, 1, - 2)$ are two vertices of $△ A B C . P, Q$ and $R$ are the midpoints of $A B$ , $B C$ and $A C$ respectively. The coordinates of $R$ are $(\frac{5}{2}, \frac{5}{2}, - \frac{7}{2})$ . Then the centroid of $△ P Q R$ is

1

(3, - 2, - 3)

2

(3, 2, - 3)

3

(3, 2, 3)

4

(\frac{3}{2}, - \frac{7}{2}, \frac{3}{2})

Explanation:

(B) : Let,
$\vec{a}, \vec{b}, \vec{c}, \vec{p}, \vec{q}, \vec{r}$ be the position vectors of $A, B, C, P, Q$ , $R$ respectively.
$\vec{a} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k},$
$\vec{b} = 4 \hat{i} + \hat{j} - 2 \hat{k},$
$\vec{r} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k}$
Consider, $\vec{c} = xi + y \hat{j} + z \hat{k}$
Then, from question -
$\frac{\vec{a} + \vec{c}}{2} = \vec{r}$
$\frac{(2 \hat{i} + 3 \hat{j} - 2 \hat{k}) + (x \hat{i} + y \hat{j} + z \hat{k})}{2} = (\frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} - \frac{7}{2} \hat{k})$
$(2 + x) \hat{i} + (3 + y) \hat{j} + (z - 2) \hat{k} = 5 \hat{i} + 5 \hat{j} - 7 \hat{k}$
On comparing both side, we get -
$x = 3, y = 2, z = - 5$
Then, $\vec{c} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$
As the centroid of the triangle formed by joining the mid points of the sides of a given triangle coincides with the centroid of the given triangle.
$∴$ Centroid
$of △ PQR = (\frac{2 + 4 + 3}{3}, \frac{3 + 1 + 2}{3}, \frac{- 2 - 2 - 5}{3}) = (3, 2, - 3)$

MHT CET-2019