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Vectors - Practice Questions (22)

Question 1: It is known that the vector ${ } _ { a } , { } _ { b }$ satisfies $| \boldsymbol { a } | = 2 , | b |...

It is known that the vector ${ } _ { a } , { } _ { b }$ satisfies $| \boldsymbol { a } | = 2 , | b | = 2$ ,if the angle between ${ } _ { a }$ and ${ } _ { b }$ is $\frac { \pi } { 3 }$ ,then $| a + b | =$ ()

  • A. A. 1
  • B. B. $\sqrt { 2 }$
  • C. C. $2 \sqrt { 3 }$
  • D. D. $\sqrt { 7 }$

Answer: C

Solution: Knowledge points] known quantity product modulus, quantity product algorithm [Analysis]Using the vector product of quantities operation, to sum the modulus of the vector can be. FORMULA_5]] , the $| a + b | = \sqrt { 4 + 2 \times 2 + 4 } = 2 \sqrt { 3 }$

Question 2: It is known that the angle between the vector ${ } ^ { a }$ and the vector ${ } ^ { b }$ is $60 ^ { ...

It is known that the angle between the vector ${ } ^ { a }$ and the vector ${ } ^ { b }$ is $60 ^ { \circ }$ ,and $| a | = b \mid = 1$ ,then the value of $| a - b |$ is ()

  • A. A. 1
  • B. B. $\frac { 3 } { 2 }$
  • C. C. 2
  • D. D. $\frac { 4 } { 3 }$

Answer: A

Solution: Knowledge Points] Find the modulus of a known product of quantities, and use the definition to find the product of vectors. [Analysis]According to the product of vectors, we can find the value of $| a - b |$.

Question 3: $a = ( 2,1 ) , b = ( x , 3 ) , a \perp b$ , then $x =$ ( )

$a = ( 2,1 ) , b = ( x , 3 ) , a \perp b$ , then $x =$ ( )

  • A. A. $\frac { 1 } { 2 }$
  • B. B. $\frac { 3 } { 2 }$
  • C. C. $- \frac { 1 } { 2 }$
  • D. D. $- \frac { 3 } { 2 }$

Answer: D

Solution: Knowledge Points] The use of vector perpendicular to find parameters, vector perpendicular to the coordinates of the expression [Analysis]According to $a \perp b$, we can get $a \cdot b = 0$, and we can find the value of ${ } ^ { x }$ by performing the coordinate operation of the quantity product. $\because a \perp b$ , ${ } ^ { x }$ $\therefore a \cdot b = 2 x + 3 = 0$ $\therefore x = - \frac { 3 } { 2 }$ .

Question 4: The vector $a , b$ is known to satisfy $| \vec { a } | = \sqrt { 3 } , | b | = 1 , | 2 a - b | = 3$ ...

The vector $a , b$ is known to satisfy $| \vec { a } | = \sqrt { 3 } , | b | = 1 , | 2 a - b | = 3$ ,then $a \cdot b =$ ( )

  • A. A. - 2
  • B. B. - 1
  • C. C. 1
  • D. D. 2

Answer: C

Solution: Knowledge point] Find the product of quantities with known modulus [Analysis]Square $| 2 a - b | = 3$ on the left and right sides at the same time, simplify and substitute the values to find $a \cdot b$. [Explanation] Because $| 2 a - b | ^ { 2 } = 4 | a | ^ { 2 } - 4 a \cdot b + | b | ^ { 2 } = 9$ , $a \cdot b$ is the product of $| 2 a - b | = 3$ and $a \cdot b$. Simplify to $4 a \cdot b = 4$ and solve to $a \cdot b = 1$.

Question 5: If $| a | = 4 , | \boldsymbol { b } | = 1 , a \cdot b = 2$ is known, the angle between $a$ and $b$ i...

If $| a | = 4 , | \boldsymbol { b } | = 1 , a \cdot b = 2$ is known, the angle between $a$ and $b$ is ( ).

  • A. A. $\frac { \pi } { 6 }$
  • B. B. $\frac { \pi } { 4 }$
  • C. C. $\frac { \pi } { 3 }$
  • D. D. $\frac { 2 \pi } { 3 }$

Answer: C

Solution: Knowledge Points] Calculation of the angle between vectors [Analysis]Using the formula for calculating the angle of a vector can be solved. [Explanation] From the known $| a | = 4 , | \boldsymbol { b } | = 1 , a \cdot b = 2$ , $| a | = 4 , | \boldsymbol { b } | = 1 , a \cdot b = 2$ , $| a | = 4 , | \boldsymbol { b } | = 1 , a \cdot b = 2$ The angle between $_ { b }$ and $_ { b }$ is ${ } _ { \theta }$ , and the cosine of $\cos \theta = \frac { a \cdot b _ { \vec { b } } } { | a | | b | } = \frac { 2 } { 4 } = \frac { 1 } { 2 }$ is Since $0 \leq \theta \leq \pi$, the cosine is $\theta = \frac { \pi } { 3 }$.

Question 6: If $\mathrm { V } ^ { \mathrm { V } A B C }$ is an equilateral triangle with side 4, then $A B \cdot...

If $\mathrm { V } ^ { \mathrm { V } A B C }$ is an equilateral triangle with side 4, then $A B \cdot B C =$ is an equilateral triangle with side 4, then $A B \cdot B C =$ is an equilateral triangle with side 4.

  • A. A. - 8
  • B. B. 8
  • C. C. $- 4 \sqrt { 3 }$
  • D. D. $- \sqrt { 3 }$

Answer: A

Solution: Knowledge Points] Use the definition to find the product of quantities of vectors. [Analysis] Use the definition of the product of vectors to solve the problem. So $A B \cdot B C = | A B | \cdot | B C | \cos \left( 180 ^ { \circ } - 60 ^ { \circ } \right) = 4 \times 4 \times \left( - \frac { 1 } { 2 } \right) = - 8$ .

Question 7: Let $a , b$ be a nonzero vector, if $( a + b ) \cdot ( a - b ) = 0$ ,then (

Let $a , b$ be a nonzero vector, if $( a + b ) \cdot ( a - b ) = 0$ ,then (

  • A. A. $a = - b$
  • B. B. $a = b$
  • C. C. $a \cdot b = 0$
  • D. D. $| a | = b \mid$

Answer: D

Solution: Knowledge Points] Operational law of the product of quantities [Analysis] Simplify the equation to get the result. [Explanation] Because $( a + b ) \cdot ( a - b ) = 0$ , $( a + b ) \cdot ( a - b ) = 0$ , $( a + b ) \cdot ( a - b ) = 0$ so $a ^ { 2 } = b ^ { 2 }$ is simplified to $| a | = b \mid$ , and

Question 8: In $\bigvee A B C$, $| A B | = | A C - A B | = | B C + A B |$, then $\vee A B C$ is ( )

In $\bigvee A B C$, $| A B | = | A C - A B | = | B C + A B |$, then $\vee A B C$ is ( )

  • A. A. equilateral triangle
  • B. B. a right angled triangle
  • C. C. obtuse triangle
  • D. D. isosceles right triangle

Answer: A

Solution: [Knowledge Point]Rules of vector subtraction [Analysis] According to the linear operation of vectors can be obtained $| A B | = | B C | = | A C | _ { \text {即可判断.} }$ [Details] $\because A C - A B = B C , B C + A B = B C - B A = A C$ , $| A B | = | A C - A B | = | B C + A B |$ $\therefore | A B | = | B C | = | A C |$ , so $\mathrm { V } A B C$ is an equilateral triangle.

Question 9: If $a , b$ is two unit vectors and $a \cdot b = - \frac { 1 } { 2 }$ , then the angle between $a$ an...

If $a , b$ is two unit vectors and $a \cdot b = - \frac { 1 } { 2 }$ , then the angle between $a$ and $b$ is ( )

  • A. A. $\frac { \pi } { 6 }$
  • B. B. $\frac { \pi } { 3 }$
  • C. C. $\frac { 2 \pi } { 3 }$
  • D. D. $\frac { 5 \pi } { 6 }$

Answer: C

Solution: Knowledge Points] Calculation of the angle between vectors [Analysis] According to the angle formula can be solved. [Detailed Explanation]If $a , b$ is two unit vectors and $a \cdot b = - \frac { 1 } { 2 }$ , then $\cos a , b = \frac { \vec { a } \cdot \vec { b } } { | a | | b | } = \frac { - \frac { 1 } { 2 } } { 1 \times 1 } = - \frac { 1 } { 2 }$ then $\cos a , b = \frac { \vec { a } \cdot \vec { b } } { | a | | b | } = \frac { - \frac { 1 } { 2 } } { 1 \times 1 } = - \frac { 1 } { 2 }$ , $a \cdot b = - \frac { 1 } { 2 }$ $\because a , b \in [ 0 , \pi ] , \therefore \overrightarrow { a , b } = \frac { 2 \pi } { 3 }$ .

Question 10: The result of simplifying $O P + P S - O Q$ is equal to

The result of simplifying $O P + P S - O Q$ is equal to

  • A. A. $S P$
  • B. B. $Q P$
  • C. C. ${ } ^ { S Q }$
  • D. D. ${ } ^ { \text {QS } }$

Answer: D

Solution: Knowledge Points] The law of vector subtraction, the law of vector addition [Analysis]According to the vector addition and subtraction algorithm step-by-step simplification $O P + P S - O Q$ [Explanation] Calculate $O P + P S$: by the triangle rule of vector addition, $O P + P S = O S$. Processing $O S - O Q$ : Vector subtraction converted to addition, i.e. $O S - O Q = O S + ( - O Q ) = O S + Q O$ Calculating $Q O + O S$ : applying the triangle rule again, $Q O + O S = Q S$ In summary, the result of the simplification is ${ } ^ { Q S }$

Question 11: If ${ } ^ { a , b }$ is two unit vectors with angle ${ } ^ { 120 ^ { \circ } }$, then $| a - 2 b | =...

If ${ } ^ { a , b }$ is two unit vectors with angle ${ } ^ { 120 ^ { \circ } }$, then $| a - 2 b | =$

  • A. A. $\sqrt { 3 }$
  • B. B. 2
  • C. C. $\sqrt { 5 }$
  • D. D. $\sqrt { 7 }$

Answer: D

Solution: Knowledge Points] The product of quantities of known modes, the calculation of the angle between vectors, the law of the product of quantities. [Analysis] Find ${ } ^ { \mathrm { a } \cdot \mathrm { b } }$ to solve the problem. [Explanation] Because $\stackrel { \mathrm { r } } { \mathrm { a } } \cdot \stackrel { \mathrm { r } } { \mathrm { b } } = \stackrel { \mathrm { r } } { \mathrm { a } } | \cdot | \stackrel { \mathrm { r } } { \mathrm { b } } \left\lvert \, \cdot \cos 120 ^ { \circ } = 1 \times 1 \times \left( - \frac { 1 } { 2 } \right) = - \frac { 1 } { 2 } \right.$ , ${ } ^ { \mathrm { a } \cdot \mathrm { b } }$ is a vector product. Therefore $| \vec { a } - 2 b | = \sqrt { | a | ^ { 2 } + | 2 b | ^ { 2 } - 2 \cdot a \cdot 2 b } = \sqrt { 1 ^ { 2 } + 2 ^ { 2 } \times 1 ^ { 2 } - 4 \times \left( - \frac { 1 } { 2 } \right) } = \sqrt { 7 }$ .

Question 12: In ${ } ^ { \vee } A B C$, $A B = 5 , B C = 3 , C A = 4$, then $A B \cdot B C =$

In ${ } ^ { \vee } A B C$, $A B = 5 , B C = 3 , C A = 4$, then $A B \cdot B C =$

  • A. A. 9
  • B. B. 18
  • C. C. - 18
  • D. D. - 9

Answer: D

Solution: Knowledge Points] Use the definition of the vector product of quantities [Analysis] Determine the shape of the triangle according to the relationship between the three sides of the triangle, and then according to the definition of the vector product of quantities to find its value can be. [Details] Because $\triangle A B C$ in $A B = 5 , B C = 3 , C A = 4$ , $A B = 5 , B C = 3 , C A = 4$ , $A B = 5 , B C = 3 , C A = 4$ is the product of the three sides of a triangle. so $B C ^ { 2 } + C A ^ { 2 } = A B ^ { 2 }$ , so $B C \perp C A$ , and $\cos \angle C B A = \frac { B C } { A B } = \frac { 3 } { 5 }$ . So $A B \cdot B C = - 3 \times 5 \times \frac { 3 } { 5 } = - 9$ .

Question 13: In the parallelogram $A B C D _ { \text{in } } , A B = 4 , A D = 3 , B E = 2 E C , D F = F C$ ,then ...

In the parallelogram $A B C D _ { \text{in } } , A B = 4 , A D = 3 , B E = 2 E C , D F = F C$ ,then $A E \square E F =$

  • A. A. - 6
  • B. B. 6
  • C. C. $- \frac { 22 } { 3 }$
  • D. D. $\frac { 22 } { 3 }$

Answer: A

Solution: Knowledge Points] The law of the product of quantities, the application of the fundamental theorem of plane vectors. [Analysis]Use the linear operation of plane vectors to represent the vector ${ } ^ { A E , E F }$, and then use the law of product of quantities of vectors to solve the problem. [Explain] $A E \cdot E F = ( A B + B E ) \cdot ( E C + C F ) = \left( A B + \frac { 2 } { 3 } A D \right) \cdot \left( \frac { 1 } { 3 } A D - \frac { 1 } { 2 } A B \right) = \frac { \overrightarrow { 2 } } { 9 } A D ^ { 2 } - \frac { 1 } { 2 } A B ^ { 2 } = - 6$ is the vector ${ } ^ { A E , E F }$.

Question 14: It is known that the angle between $| p | = 8 , | q | = 6 , p$ and $q$ is $60 ^ { \circ }$ ,then $p ...

It is known that the angle between $| p | = 8 , | q | = 6 , p$ and $q$ is $60 ^ { \circ }$ ,then $p \cdot q =$

  • A. A. $24 \sqrt { 3 }$
  • B. B. $- 24 \sqrt { 3 }$
  • C. C. 24
  • D. D. . - 24

Answer: C

Solution: [Knowledge Point]Find the product of vectors by definition. [Analysis] Use the formula for the product of quantities of vectors to find the answer. [Details] $\vec { p } \cdot q = | p | \cdot | q | \cos 60 ^ { \circ } = 8 \times 6 \times \frac { 1 } { 2 } = 24$ .

Question 15: If $| a | = 2 | b | = 2 , a \cdot b = 1$ is known, $| a + b | =$ is known.

If $| a | = 2 | b | = 2 , a \cdot b = 1$ is known, $| a + b | =$ is known.

  • A. A. 1
  • B. B. 2
  • C. C. $\sqrt { 3 }$
  • D. D. $\sqrt { 7 }$

Answer: D

Solution: Knowledge Points] To find the modulus of a known product of quantities [Analysis]Using $| \overrightarrow { a + b } | ^ { 2 } = a ^ { 2 } + b ^ { 2 } + 2 a \cdot b$ , we can find the mode. [Explanation] $| \vec { a } + \vec { b } | ^ { 2 } = a ^ { 2 } + b ^ { 2 } + 2 a \cdot b = 4 + 1 + 2 = 7$ , the Therefore $| a + b | = \sqrt { 7 }$ .

Question 16: If $| a | = 1 , ( a + b ) \cdot b = 2$ is known, the range of $| b |$ is

If $| a | = 1 , ( a + b ) \cdot b = 2$ is known, the range of $| b |$ is

  • A. A. $[ 1 , + \infty )$
  • B. B. $[ 0,2 ]$
  • C. C. $[ 2 , + \infty )$
  • D. D. $[ 1,2 ]$

Answer: D

Solution: Knowledge]Knowing the product of quantities to find modulus, using the definition of the vector product of quantities [Analysis]According to the definition of the product of vectors and the nature of the operation, simplify the conditions, use the modulus to express the cosine of the angle as a function of the value of the modulus, to find the modulus range. Detailed]Know $( a + b ) \cdot b = 2$ , get $a \cdot b + b ^ { 2 } = 2$ , deform to $| \vec { a } | | b | \cos \langle a , b \rangle + | b | ^ { 2 } = 2$ , and $| \vec { a } | | b | \cos \langle a , b \rangle + | b | ^ { 2 } = 2$ , and $| \vec { a } | | b | \cos \langle a , b \rangle + | b | ^ { 2 } = 2$. Let $| b | = t ( t > 0 )$ ,then $t \cos < a , b > + t ^ { 2 } = 2$ ,and deform to $\cos \langle \vec { a } , \vec { b } \rangle = \frac { 2 - t ^ { 2 } } { t }$ , . Since $- 1 \leq \cos \langle a , b \rangle \leq 1$ ,then $- 1 \leq \frac { 2 - t ^ { 2 } } { t } \leq 1$ ,since $t > 0$ ,then $t \leq 2 - t ^ { 2 } \leq t$ ,since $t > 0$ ,then $t \leq 2 - t ^ { 2 } \leq t$ ,since $t \leq 2 - t ^ { 2 } \leq t$ Solve the set of inequalities $\left\{ \begin{array} { l } - t \leq 2 - t ^ { 2 } \\ 2 - t ^ { 2 } \leq t \end{array} \right.$ , when $t > 0$ , solve $1 \leq t \leq 2$ .

Question 17: In $\mathrm { VABC } _ { \text{in } } , A B = a , \stackrel { \mathrm { un } } { \mathrm { CB } } = ...

In $\mathrm { VABC } _ { \text{in } } , A B = a , \stackrel { \mathrm { un } } { \mathrm { CB } } = \stackrel { \prime } { b }$ ,then $C A$ is equal to

  • A. A. $a + b$
  • B. B. $b - a$
  • C. C. $a - b$
  • D. D. $- a - b$

Answer: B

Solution: Knowledge Points] The law of vector subtraction [Analysis]From the linear operation of vector can be solved. [Detailed Explanation]From the meaning of the question $C A = C B + B A = C B - A B = b - a$ .

Question 18: $( 2 a - b ) - ( a - 2 b ) =$

$( 2 a - b ) - ( a - 2 b ) =$

  • A. A. $3 a + 3 b$
  • B. B. $a - 3 b$
  • C. C. $a + b$
  • D. D. $3 a - b$

Answer: C

Solution: Knowledge Points] Mixed operations of plane vectors [Analysis]Using the linear operation of plane vectors to simplify the result. [Explanation] From the linear operation of plane vectors, we can get $( 2 a - b ) - ( a - 2 b ) = a + b$ .

Question 19: If vector $a , b$ satisfies $| a | = 2 , | b | = 2 , a \cdot b = 2$ ,then $| a - b | =$

If vector $a , b$ satisfies $| a | = 2 , | b | = 2 , a \cdot b = 2$ ,then $| a - b | =$

  • A. A. $\sqrt { 2 }$
  • B. B. 2
  • C. C. $2 \sqrt { 3 }$
  • D. D. 4

Answer: B

Solution: [Knowledge Point]Find the mode of a known quantity product [Analyze] Put $| a - b | _ { \text {平方,利用数量积的运算律即可求解.} }$ [Details] $\because | a | = 2 , | b | = 2 , a \cdot b = 2$ , $\therefore | \vec { a } - b | ^ { 2 } = ( a - b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a \cdot b = | a | ^ { 2 } + | b | ^ { 2 } - 2 a \cdot b = 4 + 4 - 4 = 4$ , $\therefore | \vec { a } - b | ^ { 2 } = ( a - b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a \cdot b = | a | ^ { 2 } + | b | ^ { 2 } - 2 a \cdot b = 4 + 4 - 4 = 4$ , $\therefore | \vec { a } - b | ^ { 2 } = ( a - b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a \cdot b = | a | ^ { 2 } + | b | ^ { 2 } - 2 a \cdot b = 4 + 4 - 4 = 4$ $\therefore | \vec { a } - b | ^ { 2 } = ( a - b ) ^ { 2 } = a ^ { 2 } + b ^ { 2 } - 2 a \cdot b = | a | ^ { 2 } + | b | ^ { 2 } - 2 a \cdot b = 4 + 4 - 4 = 4$ , $\because | a | = 2 , | b | = 2 , a \cdot b = 2$ $\therefore | a - b | = 2$ .

Question 20: It is known that the vector $a , b$ satisfies $| a | = 1 , | b | = \sqrt { 3 } , a \cdot b = 1$ ,the...

It is known that the vector $a , b$ satisfies $| a | = 1 , | b | = \sqrt { 3 } , a \cdot b = 1$ ,then $| a + b | =$

  • A. A. $\sqrt { 3 }$
  • B. B. 2
  • C. C. $\sqrt { 5 }$
  • D. D. $\sqrt { 6 }$

Answer: D

Solution: Knowledge points] known product of quantities to find the mode, the product of quantities of the operation law [Analysis]According to the meaning of the question, we can get $| \overrightarrow { a + b } | ^ { 2 } = | a | ^ { 2 } + 2 a \cdot b + | b | ^ { 2 }$, and then according to the modulus relationship operation to solve. [Analysis] $\because | \overrightarrow { a + b } | ^ { 2 } = | a | ^ { 2 } + 2 a \cdot b + | b | ^ { 2 }$ and $| a | = 1 , | b | = \sqrt { 3 } , a \cdot b = 1$, then $| a | = 1 , | b | = \sqrt { 3 } , a \cdot b = 1$, $| a | = 1 , | b | = \sqrt { 3 } , a \cdot b = 1$ and $| a | = 1 , | b | = \sqrt { 3 } , a \cdot b = 1$. Then $\left| \overrightarrow { \left. a \right| ^ { 2 } } + 2 a \cdot b + | b | ^ { 2 } = 1 + 2 + 3 \right.$ , so $| a + b \mid = \sqrt { 6 }$ .

Question 21: It is known that $| a | = 2 , | b | = 4$ and $( a + b ) \perp a$ ,then the angle between the vectors...

It is known that $| a | = 2 , | b | = 4$ and $( a + b ) \perp a$ ,then the angle between the vectors $a$ and $b$ is

  • A. A. $\frac { 2 \pi } { 3 }$
  • B. B. $\frac { \pi } { 3 }$
  • C. C. $- \frac { \pi } { 3 }$
  • D. D. $- \frac { 2 \pi } { 3 }$

Answer: A

Solution: Knowledge Points] Vector representation of perpendicular relationship, the calculation of vector angle [Analysis] $( a + b ) \cdot a = 0$ is obtained from the known conditions, combined with the properties of the operation and definition of the product of plane vectors can be found. $\cos \langle a , b \rangle _ { \text{的值,结合plane向量夹角的取值范围可得出向量 } a \text{ and } b \text { 的夹角.} }$ [Explain] Since $| a | = 2 , | b | = 4$ and $( a + b ) \perp a$ ,then $( a + b ) \cdot a = a ^ { 2 } + a \cdot b = | a | ^ { 2 } + | a | \cdot | b | \cos \langle a , b \rangle$ $= 4 + 8 \cos \langle a , b \rangle = 0$ ,so $\cos \langle \vec { a } , b \rangle = - \frac { 1 } { 2 }$ , and $\cos \langle \vec { a } , b \rangle = - \frac { 1 } { 2 }$ Since $0 \leq \langle a , b \rangle \leq \pi$ ,so $\langle \vec { a } , b \rangle = \frac { 2 \pi } { 3 }$ ,i.e. the angle between the vector $a$ and $_ { b }$ is $\frac { 2 \pi } { 3 }$ .

Question 22: It is known that the plane vector ${ } ^ { a , b }$ satisfies $| a | = 2 , a \perp ( a + b )$ ,then ...

It is known that the plane vector ${ } ^ { a , b }$ satisfies $| a | = 2 , a \perp ( a + b )$ ,then $a \cdot b =$

  • A. A. - 2
  • B. B. 2
  • C. C. - 4
  • D. D. 4

Answer: C

Solution: Knowledge Points] Operational law of product of quantities [Analysis]According to the given conditions, use the vector representation of the vertical relationship and the operator of the product of quantities to solve the problem. By $a \perp ( a + b )$, we get $a \cdot ( a + b ) = a ^ { 2 } + a \cdot b = 0$, so $a \cdot b = - a ^ { 2 } = - 4$.
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Vectors

向量

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