Let-a-i-4j-2k-b-3i-2j-7k-and-c-2i-j-4k-Find-a-vector-d-which-is-perpendicular-to-both-a-and-b-and-c-d-15-

Question Number 15352 by Tinkutara last updated on 09/Jun/17

Let \vec{a} = \overset{\land}{i} + 4 \overset{\land}{j} + 2 \overset{\land}{k}, \vec{b} = 3 \overset{\land}{i} - 2 \overset{\land}{j} + 7 \overset{\land}{k}

and \vec{c} = 2 \overset{\land}{i} - \overset{\land}{j} + 4 \overset{\land}{k} . Find a vector \vec{d}

which is perpendicular to both \vec{a} and \vec{b},

and \vec{c} \cdot \vec{d} = 15 .

Commented by prakash jain last updated on 10/Jun/17

\vec{a} \times \vec{b} = | \begin{matrix} i & j & k \\ 1 & 4 & 2 \\ 3 & - 2 & 7 \end{matrix} |

= i (28 + 4) - j (7 - 6) + k (- 2 - 12)

= 32 i - j - 14 k

\vec{a} \times \vec{b} is perpendicular to both a and b .

u n i t v e c t o r p e r p e n d i c u l a r t o a a n d b

n = \frac{a \times b}{∣ a \times b ∣} = \frac{32 i - j - 14 k}{\sqrt{32^{2} + 1 + 14^{2}}} = \frac{32 i - j - 14 k}{\sqrt{1221}}

d = k n = k \frac{32 i - j - 14 k}{\sqrt{1221}}

c \cdot d = 15

k \frac{32 \times 2 + 1 - 14 \times 4}{\sqrt{1221}} = 15

k \frac{9}{\sqrt{1221}} = 15

k = \frac{15 \sqrt{1221}}{9}

d = k n = \frac{15 \sqrt{1221}}{9} (\frac{32 i - j - 14 k}{\sqrt{1221}})

d = \frac{5}{3} (32 i - j - 14 k)

Commented by Tinkutara last updated on 10/Jun/17

But answer is \frac{1}{3} (160 \overset{\land}{i} - 5 \overset{\land}{j} + 70 \overset{\land}{k})

Commented by prakash jain last updated on 10/Jun/17

corrected . Forgot to cancel \sqrt{1221}

Commented by Tinkutara last updated on 10/Jun/17

Thanks Sir!