Let \veca=2 \hati-3 \hatj+4 \hatk, \vecb=3 \hati+4 \hatj-5 \hatk and a vector \vecc be such that \veca \times(\vecb+\vecc)+\vecb \times \vecc=\hati+8 \hatj+13 \hatk . If \veca \cdot \vecc=13 , then (24-\vecb \cdot \vecc) is equal to \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q59 with Solution

View written solutionFree

Correct answer: 46

$\vec a=(2,-3,4),\quad \vec b=(3,4,-5),\quad \vec c=(x,y,z).$

We are given

$\vec a\times(\vec b+\vec c)+\vec b\times \vec c=(1,8,13)$

and

$\vec a\cdot \vec c=13.$

We need to find

$24-\vec b\cdot \vec c.$

Using distributivity of cross product,

$\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c.$

So,

$\vec a\times\vec b+\vec a\times\vec c+\vec b\times\vec c=(1,8,13).$

Now compute

\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -3 & 4\\ 3 & 4 & -5 \end{vmatrix}.$$ Thus, $$\vec a\times\vec b=\hat i[( -3)(-5)-4\cdot 4]-\hat j[2(-5)-4\cdot 3]+\hat k[2\cdot 4-(-3)\cdot 3].$$ $$=\hat i(15-16)-\hat j(-10-12)+\hat k(8+9)$$ $$=-\hat i+22\hat j+17\hat k.$$ Hence, $$\vec a\times\vec c+\vec b\times\vec c=(1,8,13)-(-1,22,17)=(2,-14,-4).$$ Now, $$\vec a\times\vec c+\vec b\times\vec c=(\vec a+\vec b)\times \vec c.$$ Since $$\vec a+\vec b=(5,1,-1),$$ we get $$(5,1,-1)\times (x,y,z)=(2,-14,-4).$$ --- 3. **Form equations from the cross product** Compute

(5,1,-1)\times(x,y,z)= \begin{vmatrix} \hat i & \hat j & \hat k\ 5 & 1 & -1\ x & y & z \end{vmatrix}

$$=\hat i(1\cdot z-(-1)y)-\hat j(5z-(-1)x)+\hat k(5y-x).$$ So, $$(5,1,-1)\times(x,y,z)=(y+z,\ x-5z,\ 5y-x).$$ Equating components with $(2,-14,-4)$: $$y+z=2 \quad ...(1)$$ $$x-5z=-14 \quad ...(2)$$ $$5y-x=-4 \quad ...(3)$$ Also, from $\vec a\cdot \vec c=13$, $$2x-3y+4z=13 \quad ...(4)$$ --- 4. **Solve for $x,y,z$** From (2), $$x=-14+5z.$$ From (1), $$y=2-z.$$ Substitute into (3): $$5(2-z)-(-14+5z)=-4$$ $$10-5z+14-5z=-4$$ $$24-10z=-4$$ $$-10z=-28$$ $$z=\frac{14}{5}.$$ Then $$y=2-\frac{14}{5}=\frac{10-14}{5}=-\frac{4}{5},$$ $$x=-14+5\cdot \frac{14}{5}=0.$$ Check with (4): $$2(0)-3\left(-\frac45\right)+4\left(\frac{14}{5}\right)=\frac{12}{5}+\frac{56}{5}=\frac{68}{5}\neq 13.$$ This indicates an error in the sign of the $\hat j$-component while expanding the cross product. Let us correct that carefully. --- 5. **Correct cross-product expansion** For $$(5,1,-1)\times(x,y,z),$$ the components are

\big(1\cdot z-(-1)y,\ -[5z-(-1)x],\ 5y-1\cdot x\big) =(y+z,-(5z+x),5y-x).

So the correct equations are: $$y+z=2 \quad ...(1)$$ $$-(5z+x)=-14 \Rightarrow x+5z=14 \quad ...(2)$$ $$5y-x=-4 \quad ...(3)$$ And $$2x-3y+4z=13 \quad ...(4).$$ From (1), $$y=2-z.$$ From (2), $$x=14-5z.$$ Substitute into (3): $$5(2-z)-(14-5z)=-4$$ $$10-5z-14+5z=-4$$ $$-4=-4,$$ so (3) is dependent. Now use (4): $$2(14-5z)-3(2-z)+4z=13$$ $$28-10z-6+3z+4z=13$$ $$22-3z=13$$ $$3z=9$$ $$z=3.$$ Then $$y=2-3=-1,$$ $$x=14-15=-1.$$ Hence, $$\vec c=(-1,-1,3).$$ --- 6. **Compute $\vec b\cdot \vec c$** $$\vec b\cdot \vec c=(3,4,-5)\cdot(-1,-1,3)$$ $$=3(-1)+4(-1)+(-5)(3)=-3-4-15=-22.$$ Therefore, $$24-\vec b\cdot \vec c=24-(-22)=46.$$ --- 7. **Final answer** $$\boxed{46}$$