If a¯=i¯+2j¯+3k¯, b¯=-i¯+2j¯+k¯ c¯=3i¯+j¯ and d¯ is a vector normal no both a¯ and b¯ then cosc¯, d¯=

a×b=ijk123-121=i(-4)-j(4)+k(4) =4(-i-j+k) d || a×bas c, d=4(-3-1)9+1 41+1+1=-410 3=-430

If a¯=i¯+2j¯+3k¯, b¯=-i¯+2j¯+k¯ c¯=3i¯+j¯ and d¯ is a vector normal no both a¯ and b¯ then cosc¯, d¯=

a×b=ijk123-121=i(-4)-j(4)+k(4) =4(-i-j+k) d || a×bas c, d=4(-3-1)9+1 41+1+1=-410 3=-430

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If \bar{a} = \bar{i} + 2 \bar{j} + 3 \bar{k}, \bar{b} = - \bar{i} + 2 \bar{j} + \bar{k} \bar{c} = 3 \bar{i} + \bar{j} and \bar{d} is a vector normal no both \bar{a} and \bar{b} then \cos (\bar{c}, \bar{d}) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{4}{\sqrt{30}}$

$\frac{3}{\sqrt{30}}$

$\frac{2}{\sqrt{30}}$

$\frac{1}{\sqrt{30}}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} a \times b & = & |\begin{matrix} i & j & k \\ 1 & 2 & 3 \\ - 1 & 2 & 1 \end{matrix}| = i (- 4) - j (4) + k (4) \\ = & 4 (- i - j + k) \\ d | | a \times b \\ as c, d & = & \frac{4 (- 3 - 1)}{\sqrt{9 + 1} 4 \sqrt{1 + 1 + 1}} = \frac{- 4}{\sqrt{10} \sqrt{3}} = \frac{- 4}{\sqrt{30}} \end{array}$