Find the angles between the plane whose vector equations are given as \overrightarrow{r}_{1} \.(3 \stackrel{\^}{i}+4 \stackrel{\^}{k})=6 and \overrightarrow{r_{2}} \.(2 \stackrel{\^}{i}+3 \stackrel{\^}{j}-\\sqrt{3} \stackrel{\^}{k})=4.

Correct Answer is : θ=cos−1(3−2√35)

\\theta=\cos ^{-1}({3+2 \\sqrt{3}/{5})

\\theta=\cos ^{-1}({3-2 \\sqrt{3}/{5})

\\theta=\cos ^{-1}({2-3 \\sqrt{3}/{5})

\\theta=\cos ^{-1}({3-3 \\sqrt{3}/{5})

UPSC NDA Math Model Paper 5

Question : 100 of 120

Marks: +1, -0

Find the angles between the plane whose vector equations are given as

→

1.(3

)=6 and

→

.(2

−√3

)=4.

Go to Question:

Prev Question

Next Question