Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5

Q.

Let α=4i^+3j^+5k^ and β=i^+2j^4k ^.Let β1 be parallel to α and  β2 be perpendicular to α . If β=β1+β2 , then the value of 5β2.(i^+j^+k^) is 

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

11

b

7

c

6

d

9

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

detailed_solution_thumbnail

Complete Solution:

Since β1 is parallel to α, we can write it as:

β1 = λα

where λ is a scalar. To find λ, we use the projection of β onto α:

β1 = (β ⋅ α / |α|2) α

Calculate β ⋅ α:

β ⋅ α = (1)(4) + (2)(3) + (-4)(5) = 4 + 6 - 20 = -10

Calculate |α|2:

|α|2 = 42 + 32 + 52 = 16 + 9 + 25 = 50

Substitute values for β1:

β1 = (-10 / 50) α = -1/5 α

Using α = 4i + 3j + 5k:

β1 = -4/5 i - 3/5 j - k

Since β = β1 + β2, we can write:

β2 = β - β1

Substitute β = i + 2j - 4k and β1 = -4/5 i - 3/5 j - k:

β2 = (i + 2j - 4k) - (-4/5 i - 3/5 j - k)

= 9/5 i + 13/5 j - 3k

Multiply β2 by 5:

2 = 9i + 13j - 15k

Dot product with (i + j + k):

2 ⋅ (i + j + k) = (9)(1) + (13)(1) + (-15)(1)

= 9 + 13 - 15 = 7

Final Answer:

2 ⋅ (i + j + k) = 7

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon