Let a^{\to} = 6 i^{\to} - 3 j^{\to} - 6 k↖{\to} and d^{\to} = i^{\to} + j^{\to} + k^{\to} . Suppose that a^{\to} = b^{\to} + c ^{\to} where b^{\to} is parallel to d^{\to} and c^{\to} is perpendicular to d^{\to} . Then c^{\to} is -

Correct Answer is : 7i→−2j→−5k→

7 i ^{\to}- 2 j ^{\to}-5 k^{\to}

3 i^{\to} + 6 j ^{\to}- 9 k^{\to}

KVPY SX SB Exam 2015 Question Paper

Question : 17 of 120

Marks: +1, -0

Let

a↖{→} = 6 i↖{→} - 3 j↖{→} - 6 k↖{→

} and

d↖{→} = i↖{→} + j↖{→} + k↖{→}

. Suppose that

a↖{→} = b↖{→} + c ↖{→}

where

b↖{→}

is parallel to

d↖{→}

and

c↖{→}

is perpendicular to

d↖{→}

. Then

c↖{→}

is -

Solution:

b ↖{→}= λ (i↖{\^}+j↖{\^}+k↖{\^})

a↖{→} = b↖{→} + c↖{→}

c↖{→} = a↖{→}-λ(i↖{\^}+j↖{\^}+k↖{\^})

c↖{→} =(6 i↖{→} - 3 j↖{→} - 6 k↖{→ })-λ(i↖{\^}+j↖{\^}+k↖{\^})

c↖{→}=(6-λ)i↖{\^} +(-3-λ) j↖{\^} +(-6-λ)k↖{\^}

c↖{→}.λ↖{→}=6 -λ-3 -λ- 6-λ = 0

3λ=3

λ=-1

c↖{→}= 7 i ↖{\^}- 2 j ↖{\^}-5 k↖{\^}

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