© examsiri.com
Question : 68 of 160
Marks:
+1,
-0
Solution:
Given,
a=++ b=2+4−5 c=λ+2+3 b+c=(2+λ)+6−2 d= unit vector along
(b+c)=‌ =| ‌(2+λ)+6−2 |
| √(2+λ)2+36+4 |
d=‌| (2+λ)+6−2 |
| √(2+λ)2+40 |
According to the question,
d⋅a=1 ‌| [(2+λ)+6−2][++] |
| √(2+λ)2+40 |
=1 (2+λ)+6−2=√(2+λ)2+40 (2+λ)+4=(2+λ)2+40 Squaring on both sides,
(2+λ)2+42+2⋅(2+λ)4=(2+λ)2+40)
16+8(2+λ)‌‌=40 8(2+λ)‌‌=24 2+λ‌‌=3⇒λ=1
© examsiri.com
Go to Question: