## Vectors Quiz-13

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced..
Vectors Quiz-13
Q1. The edges of a parallelopiped are unit length and are parallel to non-coplanar unit vectors $\stackrel{\to }{a}$,$\stackrel{\to }{b}$,$\stackrel{\to }{c}$ such that $\stackrel{\to }{a}$$\stackrel{\to }{b}$=$\stackrel{\to }{b}$$\stackrel{\to }{c}$=$\stackrel{\to }{c}$$\stackrel{\to }{a}$=1/2 Then,the volume of the parallelopiped is
•  1/√2 cu unit
•  1/(2√2) cu unit
•  √3/2 cu unit
•  1/√3 cu unit
Solution

Q2.If the diagonals of a parallelogram are 3$\stackrel{^}{i}$+$\stackrel{^}{j}$-2$\stackrel{^}{k}$and $\stackrel{^}{i}$-3$\stackrel{^}{j}$+4$\stackrel{^}{k}$, then the lengths of its sides are
•  √8,√10
•  √6,√14
•  √5,√12
•  None of these
Solution

Q3.  The plane through the point (-1,-1,-1) and containing the line of intersection of the planes $\stackrel{\to }{r}$∙($\stackrel{^}{i}$+3$\stackrel{^}{j}$-$\stackrel{^}{k}$)=0 and $\stackrel{\to }{r}$∙($\stackrel{^}{j}$+2$\stackrel{^}{k}$)=0 is
•   $\stackrel{\to }{r}$∙($\stackrel{^}{i}$+2$\stackrel{^}{j}$-3$\stackrel{^}{k}$)=0
•  $\stackrel{\to }{r}$∙($\stackrel{^}{i}$+4$\stackrel{^}{j}$+$\stackrel{^}{k}$)=0
•  $\stackrel{\to }{r}$∙($\stackrel{^}{i}$+5$\stackrel{^}{j}$-5$\stackrel{^}{k}$)=0
•  $\stackrel{\to }{r}$∙($\stackrel{^}{i}$+$\stackrel{^}{j}$+3$\stackrel{^}{k}$)=0
Solution

Q4. Let $\stackrel{\to }{v}$=2$\stackrel{^}{i}$+$\stackrel{^}{j}$-$\stackrel{^}{k}$ and $\stackrel{\to }{w}$=$\stackrel{^}{i}$+3$\stackrel{^}{k}$,If $\stackrel{\to }{u}$ is a unit vector, then maximum value of the scalar triple product [ $\stackrel{\to }{u}$ $\stackrel{\to }{v}$ $\stackrel{\to }{w}$ ] is
•  -1
•  √10+√6
•  √59
•  √60
Solution

Q5. If $\stackrel{\to }{a}$$\stackrel{^}{i}$=$\stackrel{\to }{a}$∙($\stackrel{^}{i}$+$\stackrel{^}{j}$ )=$\stackrel{\to }{a}$∙($\stackrel{^}{i}$+$\stackrel{^}{j}$+$\stackrel{^}{k}$), then $\stackrel{\to }{a}$ is equal to
•  $\stackrel{^}{i}$
•  $\stackrel{^}{k}$
•  $\stackrel{^}{j}$
•  $\stackrel{^}{i}$+$\stackrel{^}{j}$+$\stackrel{^}{k}$
Solution

Q6. If the projection of the vector $\stackrel{\to }{a}$ on $\stackrel{\to }{b}$ is |$\stackrel{\to }{a}$×$\stackrel{\to }{b}$| and if 3$\stackrel{\to }{b}$=$\stackrel{^}{i}$+$\stackrel{^}{j}$+$\stackrel{^}{k}$, then the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is
•  π/3
•  π/2
• π/4
•  π/6
Solution

Q7.A vector which makes equal angles with the vectors 1/3 ($\stackrel{^}{i}$-2$\stackrel{^}{j}$+2$\stackrel{^}{k}$),1/5 (-4$\stackrel{^}{i}$-3$\stackrel{^}{k}$) and $\stackrel{^}{j}$ is
•  5$\stackrel{^}{i}$+$\stackrel{^}{j}$+5$\stackrel{^}{k}$
•  -5$\stackrel{^}{i}$+$\stackrel{^}{j}$+5$\stackrel{^}{k}$
•  5$\stackrel{^}{i}$-$\stackrel{^}{j}$+5$\stackrel{^}{k}$
•  5$\stackrel{^}{i}$+$\stackrel{^}{j}$-5$\stackrel{^}{k}$
Solution
(b)

Q8. If |$\stackrel{\to }{a}$ |=3,|$\stackrel{\to }{b}$ |=4 and |$\stackrel{\to }{a}$+$\stackrel{\to }{b}$ |=5, then |$\stackrel{\to }{a}$-$\stackrel{\to }{b}$ |=
•  6
•  5
•  4
•  3
Solution
(b)

Q9. If |$\stackrel{\to }{a}$ |=7,|$\stackrel{\to }{b}$ |=11,|$\stackrel{\to }{a}$+$\stackrel{\to }{b}$ |=10√3, then |$\stackrel{\to }{a}$-$\stackrel{\to }{b}$ | equals
•  10
•  √10
•  2√10
•  20
Solution

Q10. If $\stackrel{\to }{a}$,$\stackrel{\to }{b}$ and $\stackrel{\to }{c}$ are position vectors of the vertices of the triangle ABC, then |($\stackrel{\to }{a}$-$\stackrel{\to }{c}$ )×($\stackrel{\to }{b}$-$\stackrel{\to }{a}$ )|/(($\stackrel{\to }{c}$-$\stackrel{\to }{a}$ )∙($\stackrel{\to }{b}$-$\stackrel{\to }{a}$ ) ) is equal to
•   cot⁡A
•   cot⁡C
•  -tan⁡C
•   tan⁡A
Solution

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