QBManager

Vector Algebra

87828 If $a, b, c$ are unit vectors satisfying the relation $a + b + \sqrt{3} c = 0$ , then the angle between $a$ and $b$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{3}

4

\frac{π}{2}

Explanation:

(C) : Given, $\vec{a} + \vec{b} = - \sqrt{3} \vec{c}$
On squaring above equation, we get-
$| \vec{a} |^{2} + | \vec{b} |^{2} + 2 | \vec{a} | | \vec{b} | \cos θ = 3 | \vec{c} |^{2}$
$1 + 1 + 2 (1) (1) \cos θ = 3 (1) [∵ | \vec{a} | = | \vec{b} | = | \vec{c} | = 1]$
$\cos θ = \frac{1}{2}$
$θ = \frac{π}{3}$

TS EAMCET-2016

Vector Algebra

87810 If vectors $a_{1} = x \hat{i} - \hat{j} + \hat{k}$ and $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is

1

\frac{1}{\sqrt{2}} (- \hat{j} + \hat{k})

2

\frac{1}{\sqrt{2}} (\hat{i} - \hat{j})

3

\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} - \hat{k})

4

\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})

Explanation:

(D) : Given that, $a_{1} = x \hat{i} - \hat{j} + \hat{k}$
And, $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear,
Then, $\frac{x}{1} = \frac{- 1}{y} = \frac{1}{z} = λ$ (say)
This gives $x = λ, y = \frac{- 1}{λ}, z = \frac{1}{λ}$
Then, unit vector parallel to vector $x \hat{i} + y \hat{j} + z \hat{k}$ will be $= \frac{((λ) \hat{i} - (\frac{1}{λ}) \hat{j} + (\frac{1}{λ}) \hat{k})}{\sqrt{(λ)^{2} + {(\frac{- 1}{λ})}^{2} + {(\frac{1}{λ})}^{2}}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k}) λ}{λ \sqrt{λ^{4} + 2}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k})}{\sqrt{λ^{4} + 2}}$ Take, $λ = 1$
Then, $x \hat{i} + \hat{y} + z \hat{k} = \frac{(\hat{i} - \hat{j} + \hat{k})}{\sqrt{3}}$

JEE Main-2021-26.02.2021

Vector Algebra

87811 The magnitude of the projection of the vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + 3 \hat{k}$ is

1

\frac{\sqrt{3}}{2}

2

\sqrt{6}

3

3 \sqrt{6}

4

\sqrt{\frac{3}{2}}

Explanation:

(D) : According to the question,
The vector perpendicular to the given vector is,
$\vec{a} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array} |$
$\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$
$Projection of \vec{b} on \vec{a} = \frac{| \vec{a} \cdot \vec{b} |}{| \vec{a} |}$
$= \frac{| (\hat{i} - 2 \hat{j} + \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k}) |}{\sqrt{6}} = \frac{| 2 - 6 + 1 |}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}}$

JEE Main-2019-08.04.2019-Shift I

Vector Algebra

87812 If $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$ and $\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ then the magnitude of the projection on $\vec{c}$ of $\vec{a}$ unit vector that is perpendicular to both $\vec{a}$ and $\vec{b}$ is

1

\frac{1}{\sqrt{29} \sqrt{3}}

2

\frac{1}{\sqrt{6}}

3

\frac{1}{\sqrt{58}}

4

\frac{3}{\sqrt{29}}

Explanation:

(C) : Given that,
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$
$\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$
Then,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array} |$
$\hat{i} (2 - 1) - \hat{j} (2 - 1) + \hat{k} (1 - 1) = \hat{i} - \hat{j}$
So, unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is,
$= \pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$
Now, the magnitude of the projection of $\pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$ on $\vec{c}$ is,
$= | \pm (\frac{\hat{i} - \hat{j}}{\sqrt{2}}) \cdot \frac{(2 \hat{i} + 3 \hat{j} - 4 \hat{k})}{\sqrt{4 + 9 + 16}} | = | \frac{2 - 3}{\sqrt{2} \sqrt{29}} | = \frac{1}{\sqrt{58}}$

AP EAMCET-2019-23.04.2019

Vector Algebra

87828 If $a, b, c$ are unit vectors satisfying the relation $a + b + \sqrt{3} c = 0$ , then the angle between $a$ and $b$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{3}

4

\frac{π}{2}

Explanation:

(C) : Given, $\vec{a} + \vec{b} = - \sqrt{3} \vec{c}$
On squaring above equation, we get-
$| \vec{a} |^{2} + | \vec{b} |^{2} + 2 | \vec{a} | | \vec{b} | \cos θ = 3 | \vec{c} |^{2}$
$1 + 1 + 2 (1) (1) \cos θ = 3 (1) [∵ | \vec{a} | = | \vec{b} | = | \vec{c} | = 1]$
$\cos θ = \frac{1}{2}$
$θ = \frac{π}{3}$

TS EAMCET-2016

Vector Algebra

87810 If vectors $a_{1} = x \hat{i} - \hat{j} + \hat{k}$ and $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is

1

\frac{1}{\sqrt{2}} (- \hat{j} + \hat{k})

2

\frac{1}{\sqrt{2}} (\hat{i} - \hat{j})

3

\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} - \hat{k})

4

\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})

Explanation:

(D) : Given that, $a_{1} = x \hat{i} - \hat{j} + \hat{k}$
And, $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear,
Then, $\frac{x}{1} = \frac{- 1}{y} = \frac{1}{z} = λ$ (say)
This gives $x = λ, y = \frac{- 1}{λ}, z = \frac{1}{λ}$
Then, unit vector parallel to vector $x \hat{i} + y \hat{j} + z \hat{k}$ will be $= \frac{((λ) \hat{i} - (\frac{1}{λ}) \hat{j} + (\frac{1}{λ}) \hat{k})}{\sqrt{(λ)^{2} + {(\frac{- 1}{λ})}^{2} + {(\frac{1}{λ})}^{2}}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k}) λ}{λ \sqrt{λ^{4} + 2}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k})}{\sqrt{λ^{4} + 2}}$ Take, $λ = 1$
Then, $x \hat{i} + \hat{y} + z \hat{k} = \frac{(\hat{i} - \hat{j} + \hat{k})}{\sqrt{3}}$

JEE Main-2021-26.02.2021

Vector Algebra

87811 The magnitude of the projection of the vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + 3 \hat{k}$ is

1

\frac{\sqrt{3}}{2}

2

\sqrt{6}

3

3 \sqrt{6}

4

\sqrt{\frac{3}{2}}

Explanation:

(D) : According to the question,
The vector perpendicular to the given vector is,
$\vec{a} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array} |$
$\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$
$Projection of \vec{b} on \vec{a} = \frac{| \vec{a} \cdot \vec{b} |}{| \vec{a} |}$
$= \frac{| (\hat{i} - 2 \hat{j} + \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k}) |}{\sqrt{6}} = \frac{| 2 - 6 + 1 |}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}}$

JEE Main-2019-08.04.2019-Shift I

Vector Algebra

87812 If $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$ and $\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ then the magnitude of the projection on $\vec{c}$ of $\vec{a}$ unit vector that is perpendicular to both $\vec{a}$ and $\vec{b}$ is

1

\frac{1}{\sqrt{29} \sqrt{3}}

2

\frac{1}{\sqrt{6}}

3

\frac{1}{\sqrt{58}}

4

\frac{3}{\sqrt{29}}

Explanation:

(C) : Given that,
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$
$\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$
Then,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array} |$
$\hat{i} (2 - 1) - \hat{j} (2 - 1) + \hat{k} (1 - 1) = \hat{i} - \hat{j}$
So, unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is,
$= \pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$
Now, the magnitude of the projection of $\pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$ on $\vec{c}$ is,
$= | \pm (\frac{\hat{i} - \hat{j}}{\sqrt{2}}) \cdot \frac{(2 \hat{i} + 3 \hat{j} - 4 \hat{k})}{\sqrt{4 + 9 + 16}} | = | \frac{2 - 3}{\sqrt{2} \sqrt{29}} | = \frac{1}{\sqrt{58}}$

AP EAMCET-2019-23.04.2019

Vector Algebra

87828 If $a, b, c$ are unit vectors satisfying the relation $a + b + \sqrt{3} c = 0$ , then the angle between $a$ and $b$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{3}

4

\frac{π}{2}

Explanation:

(C) : Given, $\vec{a} + \vec{b} = - \sqrt{3} \vec{c}$
On squaring above equation, we get-
$| \vec{a} |^{2} + | \vec{b} |^{2} + 2 | \vec{a} | | \vec{b} | \cos θ = 3 | \vec{c} |^{2}$
$1 + 1 + 2 (1) (1) \cos θ = 3 (1) [∵ | \vec{a} | = | \vec{b} | = | \vec{c} | = 1]$
$\cos θ = \frac{1}{2}$
$θ = \frac{π}{3}$

TS EAMCET-2016

Vector Algebra

87810 If vectors $a_{1} = x \hat{i} - \hat{j} + \hat{k}$ and $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is

1

\frac{1}{\sqrt{2}} (- \hat{j} + \hat{k})

2

\frac{1}{\sqrt{2}} (\hat{i} - \hat{j})

3

\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} - \hat{k})

4

\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})

Explanation:

(D) : Given that, $a_{1} = x \hat{i} - \hat{j} + \hat{k}$
And, $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear,
Then, $\frac{x}{1} = \frac{- 1}{y} = \frac{1}{z} = λ$ (say)
This gives $x = λ, y = \frac{- 1}{λ}, z = \frac{1}{λ}$
Then, unit vector parallel to vector $x \hat{i} + y \hat{j} + z \hat{k}$ will be $= \frac{((λ) \hat{i} - (\frac{1}{λ}) \hat{j} + (\frac{1}{λ}) \hat{k})}{\sqrt{(λ)^{2} + {(\frac{- 1}{λ})}^{2} + {(\frac{1}{λ})}^{2}}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k}) λ}{λ \sqrt{λ^{4} + 2}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k})}{\sqrt{λ^{4} + 2}}$ Take, $λ = 1$
Then, $x \hat{i} + \hat{y} + z \hat{k} = \frac{(\hat{i} - \hat{j} + \hat{k})}{\sqrt{3}}$

JEE Main-2021-26.02.2021

Vector Algebra

87811 The magnitude of the projection of the vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + 3 \hat{k}$ is

1

\frac{\sqrt{3}}{2}

2

\sqrt{6}

3

3 \sqrt{6}

4

\sqrt{\frac{3}{2}}

Explanation:

(D) : According to the question,
The vector perpendicular to the given vector is,
$\vec{a} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array} |$
$\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$
$Projection of \vec{b} on \vec{a} = \frac{| \vec{a} \cdot \vec{b} |}{| \vec{a} |}$
$= \frac{| (\hat{i} - 2 \hat{j} + \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k}) |}{\sqrt{6}} = \frac{| 2 - 6 + 1 |}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}}$

JEE Main-2019-08.04.2019-Shift I

Vector Algebra

87812 If $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$ and $\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ then the magnitude of the projection on $\vec{c}$ of $\vec{a}$ unit vector that is perpendicular to both $\vec{a}$ and $\vec{b}$ is

1

\frac{1}{\sqrt{29} \sqrt{3}}

2

\frac{1}{\sqrt{6}}

3

\frac{1}{\sqrt{58}}

4

\frac{3}{\sqrt{29}}

Explanation:

(C) : Given that,
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$
$\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$
Then,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array} |$
$\hat{i} (2 - 1) - \hat{j} (2 - 1) + \hat{k} (1 - 1) = \hat{i} - \hat{j}$
So, unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is,
$= \pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$
Now, the magnitude of the projection of $\pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$ on $\vec{c}$ is,
$= | \pm (\frac{\hat{i} - \hat{j}}{\sqrt{2}}) \cdot \frac{(2 \hat{i} + 3 \hat{j} - 4 \hat{k})}{\sqrt{4 + 9 + 16}} | = | \frac{2 - 3}{\sqrt{2} \sqrt{29}} | = \frac{1}{\sqrt{58}}$

AP EAMCET-2019-23.04.2019

Vector Algebra

87828 If $a, b, c$ are unit vectors satisfying the relation $a + b + \sqrt{3} c = 0$ , then the angle between $a$ and $b$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{3}

4

\frac{π}{2}

Explanation:

(C) : Given, $\vec{a} + \vec{b} = - \sqrt{3} \vec{c}$
On squaring above equation, we get-
$| \vec{a} |^{2} + | \vec{b} |^{2} + 2 | \vec{a} | | \vec{b} | \cos θ = 3 | \vec{c} |^{2}$
$1 + 1 + 2 (1) (1) \cos θ = 3 (1) [∵ | \vec{a} | = | \vec{b} | = | \vec{c} | = 1]$
$\cos θ = \frac{1}{2}$
$θ = \frac{π}{3}$

TS EAMCET-2016

Vector Algebra

87810 If vectors $a_{1} = x \hat{i} - \hat{j} + \hat{k}$ and $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is

1

\frac{1}{\sqrt{2}} (- \hat{j} + \hat{k})

2

\frac{1}{\sqrt{2}} (\hat{i} - \hat{j})

3

\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} - \hat{k})

4

\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})

Explanation:

(D) : Given that, $a_{1} = x \hat{i} - \hat{j} + \hat{k}$
And, $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear,
Then, $\frac{x}{1} = \frac{- 1}{y} = \frac{1}{z} = λ$ (say)
This gives $x = λ, y = \frac{- 1}{λ}, z = \frac{1}{λ}$
Then, unit vector parallel to vector $x \hat{i} + y \hat{j} + z \hat{k}$ will be $= \frac{((λ) \hat{i} - (\frac{1}{λ}) \hat{j} + (\frac{1}{λ}) \hat{k})}{\sqrt{(λ)^{2} + {(\frac{- 1}{λ})}^{2} + {(\frac{1}{λ})}^{2}}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k}) λ}{λ \sqrt{λ^{4} + 2}} = \frac{(λ^{2} \hat{i} - \hat{j} + \hat{k})}{\sqrt{λ^{4} + 2}}$ Take, $λ = 1$
Then, $x \hat{i} + \hat{y} + z \hat{k} = \frac{(\hat{i} - \hat{j} + \hat{k})}{\sqrt{3}}$

JEE Main-2021-26.02.2021

Vector Algebra

87811 The magnitude of the projection of the vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + 3 \hat{k}$ is

1

\frac{\sqrt{3}}{2}

2

\sqrt{6}

3

3 \sqrt{6}

4

\sqrt{\frac{3}{2}}

Explanation:

(D) : According to the question,
The vector perpendicular to the given vector is,
$\vec{a} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array} |$
$\vec{a} = \hat{i} - 2 \hat{j} + \hat{k}$
$Projection of \vec{b} on \vec{a} = \frac{| \vec{a} \cdot \vec{b} |}{| \vec{a} |}$
$= \frac{| (\hat{i} - 2 \hat{j} + \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k}) |}{\sqrt{6}} = \frac{| 2 - 6 + 1 |}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}}$

JEE Main-2019-08.04.2019-Shift I

Vector Algebra

87812 If $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$ and $\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$ then the magnitude of the projection on $\vec{c}$ of $\vec{a}$ unit vector that is perpendicular to both $\vec{a}$ and $\vec{b}$ is

1

\frac{1}{\sqrt{29} \sqrt{3}}

2

\frac{1}{\sqrt{6}}

3

\frac{1}{\sqrt{58}}

4

\frac{3}{\sqrt{29}}

Explanation:

(C) : Given that,
$\vec{a} = \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + \hat{j} + 2 \hat{k}$
$\vec{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$
Then,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array} |$
$\hat{i} (2 - 1) - \hat{j} (2 - 1) + \hat{k} (1 - 1) = \hat{i} - \hat{j}$
So, unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is,
$= \pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$
Now, the magnitude of the projection of $\pm \frac{(\hat{i} - \hat{j})}{\sqrt{2}}$ on $\vec{c}$ is,
$= | \pm (\frac{\hat{i} - \hat{j}}{\sqrt{2}}) \cdot \frac{(2 \hat{i} + 3 \hat{j} - 4 \hat{k})}{\sqrt{4 + 9 + 16}} | = | \frac{2 - 3}{\sqrt{2} \sqrt{29}} | = \frac{1}{\sqrt{58}}$

AP EAMCET-2019-23.04.2019

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