Solveeit Logo
Login

Question

Mathematics Question on Vector Algebra

If a⃗=2i^+3j^−k^,b⃗=i^+2j^−5k^,c⃗=3i^+5j^−k^,\vec {a}= {2\hat{i}}+3\hat{j}-\hat k,\vec b = \hat i+2 \hat j-5 \hat k ,\vec{ c} =3 \hat i+ 5\hat j-\hat k, then a vector perpendicular to a⃗\vec{a} and in the plane containing b⃗\vec {b} and c⃗\vec {c} is

A

17i^+21j^−123k^17 \hat i+21 \hat{j} -123 \hat k

B

−17i^+21j^−97k^-17 \hat i+21 \hat{j} -97 \hat k

C

−17i^−21j^−97k^-17 \hat i-21 \hat{j} -97 \hat k

D

−17i^−21j^+97k^-17 \hat i-21 \hat{j} +97 \hat k

Answer

−17i^−21j^−97k^-17 \hat i-21 \hat{j} -97 \hat k

Explanation

Solution

We know that a vector perpendicular to a and in the plane containing b⃗\vec{b} and c⃗\vec{c} is given by
a⃗×(b⃗×c⃗).\vec{a} \times (\vec{b} \times \vec{c}).
Given a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3 \hat{j} - \hat{k}
b⃗=i^+2j^−5k^\vec{b} = \hat{i} + 2 \hat{j} - 5 \hat{k} and c⃗=3i^+5j^−k^\vec{c} = 3 \hat{i } + 5 \hat{j} - \hat{k}
∴b⃗×c⃗=∣i^j^k^ 12−5 35−1∣=23i^−14j^−k^\therefore \vec{b} \times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 1 &2&-5\\\ 3&5&-1\end{vmatrix} = 23 \hat{i} - 14 \hat{j} - \hat{k}
Now, a⃗×(b⃗×c⃗)=∣i^j^k^ 23−1 23−14−1∣\vec{a} \times\left( \vec{b} \times\vec{c}\right) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 2&3&-1\\\ 23&-14&-1\end{vmatrix}
=(−3−14)i^−j^(−2+23)+k^(−28−69)= (-3 - 14) \hat{ i} - \hat{j} (-2+ 23) + \hat{k} (-28 - 69)
=−17i^−21j^−97k^= -17 \hat{i} - 21 \hat{ j} - 9 7 \hat{k}
Which is the required vector.