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Exercise 6.3 · Q4

Q.If a⃗=2i^+3j^−k^, b⃗=3i^+5j^+2k^, c⃗=−i^−2j^+3k^\vec a=2\hat i+3\hat j-\hat k,\ \vec b=3\hat i+5\hat j+2\hat k,\ \vec c=-\hat i-2\hat j+3\hat k, verify that

(i) (a⃗×b⃗)×c⃗=(a⃗⋅c⃗)b⃗−(b⃗⋅c⃗)a⃗(\vec a\times\vec b)\times\vec c=(\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a
(ii) a⃗×(b⃗×c⃗)=(a⃗⋅c⃗)b⃗−(a⃗⋅b⃗)c⃗\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c.
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Compute the LEFT side of each identity directly via cross products, then compute the RIGHT side via the two dot products a⃗⋅c⃗,b⃗⋅c⃗\vec a\cdot\vec c,\vec b\cdot\vec c (or a⃗⋅c⃗,a⃗⋅b⃗\vec a\cdot\vec c,\vec a\cdot\vec b) — both routes must agree.

Step 1. Useful dot products. a⃗⋅b⃗=(2)(3)+(3)(5)+(−1)(2)=6+15−2=19\vec a\cdot\vec b=(2)(3)+(3)(5)+(-1)(2)=6+15-2=19; a⃗⋅c⃗=(2)(−1)+(3)(−2)+(−1)(3)=−2−6−3=−11\vec a\cdot\vec c=(2)(-1)+(3)(-2)+(-1)(3)=-2-6-3=-11; b⃗⋅c⃗=(3)(−1)+(5)(−2)+(2)(3)=−3−10+6=−7\vec b\cdot\vec c=(3)(-1)+(5)(-2)+(2)(3)=-3-10+6=-7.

Step 2. (i) LHS: compute a⃗×b⃗\vec a\times\vec b then cross with c⃗\vec c.

a⃗×b⃗=∣i^j^k^23−1352∣=i^(6+5)−j^(4+3)+k^(10−9)=11i^−7j^+k^.\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&-1\\3&5&2\end{vmatrix}=\hat i(6+5)-\hat j(4+3)+\hat k(10-9)=11\hat i-7\hat j+\hat k.

(a⃗×b⃗)×c⃗=∣i^j^k^11−71−1−23∣=i^(−21+2)−j^(33+1)+k^(−22−7)=−19i^−34j^−29k^.(\vec a\times\vec b)\times\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\11&-7&1\\-1&-2&3\end{vmatrix}=\hat i(-21+2)-\hat j(33+1)+\hat k(-22-7)=-19\hat i-34\hat j-29\hat k.

Step 3. (i) RHS: (a⃗⋅c⃗)b⃗−(b⃗⋅c⃗)a⃗(\vec a\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec a.

=−11b⃗−(−7)a⃗=−11(3i^+5j^+2k^)+7(2i^+3j^−k^)=(−33i−55j−22k)+(14i+21j−7k)=−19i^−34j^−29k^.=-11\vec b-(-7)\vec a=-11(3\hat i+5\hat j+2\hat k)+7(2\hat i+3\hat j-\hat k)=(-33i-55j-22k)+(14i+21j-7k)=-19\hat i-34\hat j-29\hat k.

LHS == RHS. Identity (i) verified.

Step 4. (ii) LHS: compute b⃗×c⃗\vec b\times\vec c then cross with a⃗\vec a.

b⃗×c⃗=∣i^j^k^352−1−23∣=i^(15+4)−j^(9+2)+k^(−6+5)=19i^−11j^−k^.\vec b\times\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\3&5&2\\-1&-2&3\end{vmatrix}=\hat i(15+4)-\hat j(9+2)+\hat k(-6+5)=19\hat i-11\hat j-\hat k. …

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