Let \veca=3 \hati+2 \hatj+\hatk, \vecb=2 \hati-\hatj+3 \hatk and \vecc be a vector such that (\veca+\vecb) \times \vecc=2(\veca \times \vecb)+24 \hatj-6 \hatk and (\veca-\vecb+\hati) \cdot \vecc=-3 . Then |\vecc|^2 is equal to \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q51 with Solution

View written solutionFree

Correct answer: 38

Given vectors

$\vec a=(3,2,1),\qquad \vec b=(2,-1,3)$

Let

$\vec c=(x,y,z).$

We need to find $|\vec c|^2=x^2+y^2+z^2$ .

Compute $\vec a+\vec b$ and $\vec a\times \vec b$

First,

$\vec a+\vec b=(3+2,\,2+(-1),\,1+3)=(5,1,4).$

Now,

\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 2 & 1\\ 2 & -1 & 3 \end{vmatrix}$$ $$=\hat i(2\cdot 3-1\cdot(-1)) - \hat j(3\cdot 3-1\cdot 2)+\hat k(3\cdot(-1)-2\cdot 2)$$ $$=7\hat i-7\hat j-7\hat k.$$ Hence, $$2(\vec a\times \vec b)=14\hat i-14\hat j-14\hat k.$$ So the first condition becomes $$ (\vec a+\vec b)\times \vec c = 2(\vec a\times \vec b)+24\hat j-6\hat k $$ $$ (5,1,4)\times (x,y,z)=(14,-14+24,-14-6)=(14,10,-20). $$ --- 3. **Expand the cross product** $$ (5,1,4)\times (x,y,z) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 5 & 1 & 4\\ x & y & z \end{vmatrix}$$ $$=\hat i(z-4y)-\hat j(5z-4x)+\hat k(5y-x).$$ So, $$ (z-4y,\,-5z+4x,\,5y-x)=(14,10,-20). $$ Thus we get: $$z-4y=14 \qquad ...(1)$$ $$-5z+4x=10 \qquad ...(2)$$ $$5y-x=-20 \qquad ...(3)$$ --- 4. **Use the dot product condition** Given: $$(\vec a-\vec b+\hat i)\cdot \vec c=-3.$$ First, $$\vec a-\vec b=(3-2,\,2-(-1),\,1-3)=(1,3,-2).$$ So, $$\vec a-\vec b+\hat i=(2,3,-2).$$ Therefore, $$2x+3y-2z=-3. \qquad ...(4)$$ --- 5. **Solve for $x,y,z$** From (3): $$x=5y+20. \qquad ...(5)$$ From (1): $$z=14+4y. \qquad ...(6)$$ Substitute (5) and (6) into (4): $$2(5y+20)+3y-2(14+4y)=-3$$ $$10y+40+3y-28-8y=-3$$ $$5y+12=-3$$ $$5y=-15$$ $$y=-3.$$ Then, $$x=5(-3)+20=5,$$ $$z=14+4(-3)=2.$$ So, $$\vec c=(5,-3,2).$$ Check in (2): $$-5(2)+4(5)=-10+20=10,$$ which is correct. --- 6. **Find $|\vec c|^2$** $$|\vec c|^2=5^2+(-3)^2+2^2=25+9+4=38.$$ --- 7. **Comparison with stored answer** Derived answer is **38**, which matches the stored correct answer **38**.

More from Vector Algebra

$A (2, 6, 2), B (- 4, 0, λ), C (2, 3, - 1)$ and $D (4, 5, 0), ∣ λ ∣ \leq 5$ are the vertices of a quadrilateral $A B C D$ . If its area is 18 square units, then $5 - 6 λ$ is equal to .2023 · Numerical
Let $a = 5 \hat{i} - \hat{j} - 3 \hat{k}$ and $b = \hat{i} + 3 \hat{j} + 5 \hat{k}$ be two vectors. Then which one of the following statements is TRUE ?2023 · MCQ
Let $a = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}, b = \hat{i} + \hat{k}$ and $c = \hat{i} + 2 \hat{j} - 3 \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ …2023 · MCQ
Let $a = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, b = \hat{i} - 2 \hat{j} - 2 \hat{k}$ and $c = - \hat{i} + 4 \hat{j} + 3 \hat{k}$ . If $d$ is a vector perpendicular to both $b$ and $c$ , and $a \cdot d = 18$ , then $∣ a \times d ∣^{2}$ …2023 · MCQ
If the points with position vectors $α \hat{i} + 10 \hat{j} + 13 \hat{k}, 6 \hat{i} + 11 \hat{j} + 11 \hat{k}, \frac{9}{2} \hat{i} + β \hat{j} - 8 \hat{k}$ are collinear, then $(19 α - 6 β)^{2}$ is equal to :2023 · MCQ
Let $a = 6 \hat{i} + 9 \hat{j} + 12 \hat{k}, b = α \hat{i} + 11 \hat{j} - 2 \hat{k}$ and $c$ be vectors such that $a \times c = a \times b$ . If $a \cdot c = - 12, c \cdot (\hat{i} - 2 \hat{j} + \hat{k}) = 5$ …2023 · Numerical
The area of the quadrilateral $ABCD$ with vertices $A (2, 1, 1), B (1, 2, 5), C (- 2, - 3, 5)$ and $D (1, - 6, - 7)$ is equal to :2023 · MCQ
Let O be the origin and the position vector of the point P be $- i - 2 j + 3 k$ . If the position vectors of the points A, B and C are $- 2 i + j - 3 k, 2 i + 4 j - 2 k$ …2023 · MCQ

View written solutionFree

Correct answer: 38

Given vectors

$\vec a=(3,2,1),\qquad \vec b=(2,-1,3)$

Let

$\vec c=(x,y,z).$

We need to find $|\vec c|^2=x^2+y^2+z^2$ .

Compute $\vec a+\vec b$ and $\vec a\times \vec b$

First,

$\vec a+\vec b=(3+2,\,2+(-1),\,1+3)=(5,1,4).$

Now,

\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 2 & 1\\ 2 & -1 & 3 \end{vmatrix}$$ $$=\hat i(2\cdot 3-1\cdot(-1)) - \hat j(3\cdot 3-1\cdot 2)+\hat k(3\cdot(-1)-2\cdot 2)$$ $$=7\hat i-7\hat j-7\hat k.$$ Hence, $$2(\vec a\times \vec b)=14\hat i-14\hat j-14\hat k.$$ So the first condition becomes $$ (\vec a+\vec b)\times \vec c = 2(\vec a\times \vec b)+24\hat j-6\hat k $$ $$ (5,1,4)\times (x,y,z)=(14,-14+24,-14-6)=(14,10,-20). $$ --- 3. **Expand the cross product** $$ (5,1,4)\times (x,y,z) =\begin{vmatrix} \hat i & \hat j & \hat k\\ 5 & 1 & 4\\ x & y & z \end{vmatrix}$$ $$=\hat i(z-4y)-\hat j(5z-4x)+\hat k(5y-x).$$ So, $$ (z-4y,\,-5z+4x,\,5y-x)=(14,10,-20). $$ Thus we get: $$z-4y=14 \qquad ...(1)$$ $$-5z+4x=10 \qquad ...(2)$$ $$5y-x=-20 \qquad ...(3)$$ --- 4. **Use the dot product condition** Given: $$(\vec a-\vec b+\hat i)\cdot \vec c=-3.$$ First, $$\vec a-\vec b=(3-2,\,2-(-1),\,1-3)=(1,3,-2).$$ So, $$\vec a-\vec b+\hat i=(2,3,-2).$$ Therefore, $$2x+3y-2z=-3. \qquad ...(4)$$ --- 5. **Solve for $x,y,z$** From (3): $$x=5y+20. \qquad ...(5)$$ From (1): $$z=14+4y. \qquad ...(6)$$ Substitute (5) and (6) into (4): $$2(5y+20)+3y-2(14+4y)=-3$$ $$10y+40+3y-28-8y=-3$$ $$5y+12=-3$$ $$5y=-15$$ $$y=-3.$$ Then, $$x=5(-3)+20=5,$$ $$z=14+4(-3)=2.$$ So, $$\vec c=(5,-3,2).$$ Check in (2): $$-5(2)+4(5)=-10+20=10,$$ which is correct. --- 6. **Find $|\vec c|^2$** $$|\vec c|^2=5^2+(-3)^2+2^2=25+9+4=38.$$ --- 7. **Comparison with stored answer** Derived answer is **38**, which matches the stored correct answer **38**.

Vector Algebra question

Vector Algebra question

More from Vector Algebra

Vector Algebra question

Vector Algebra question

More from Vector Algebra