Let \overrightarrow u , \overrightarrow v and \overrightarrow w be vectors in three-dimensional space, where \overrightarrow u and \overrightarrow v are unit vectors which are not perpendicular to each other and \overrightarrow u . \overrightarrow w = 1, \overrightarrow v . \overrightarrow w = 1, \overrightarrow w . \overrightarrow w = 4 If the volume of the paralleopiped, whose adjacent sides are represented by the vectors, \overrightarrow u , \overrightarrow v and \overrightarrow w , is \sqrt 2 , then the value of | 3\overrightarrow u + 5\overrightarrow v | is \underline\hspace2cm .

Vector Algebra PYQ — JEE Advanced 2021 · Shift 1 · Q38 with Solution

View written solutionFree

Correct answer: 7

Given data

Let $|\,\vec u\,|=1,\qquad |\,\vec v\,|=1$ and $\vec u,\vec v$ are not perpendicular.

Also, $\vec u\cdot \vec w=1,\qquad \vec v\cdot \vec w=1,\qquad \vec w\cdot \vec w=4$ so $|\vec w|^2=4 \implies |\vec w|=2.$

The volume of the parallelepiped formed by $\vec u,\vec v,\vec w$ is $|\vec u\cdot(\vec v\times \vec w)|=\sqrt2.$

We need to find $|3\vec u+5\vec v|.$

Use Gram determinant for volume

For vectors $\vec u,\vec v,\vec w$ , the square of the volume is the determinant of the Gram matrix:

\vec u\cdot\vec u & \vec u\cdot\vec v & \vec u\cdot\vec w\\ \vec v\cdot\vec u & \vec v\cdot\vec v & \vec v\cdot\vec w\\ \vec w\cdot\vec u & \vec w\cdot\vec v & \vec w\cdot\vec w \end{pmatrix}.$$ Let $$\vec u\cdot\vec v=x.$$ Then the Gram matrix becomes $$\begin{pmatrix} 1 & x & 1\\ x & 1 & 1\\ 1 & 1 & 4 \end{pmatrix}.$$ Since $V=\sqrt2$, we have $$V^2=2.$$ So, $$\det\begin{pmatrix} 1 & x & 1\\ x & 1 & 1\\ 1 & 1 & 4 \end{pmatrix}=2.$$ --- 3. **Compute the determinant** Expand along the first row: $$\det = 1\begin{vmatrix}1&1\\1&4\end{vmatrix}-x\begin{vmatrix}x&1\\1&4\end{vmatrix}+1\begin{vmatrix}x&1\\1&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}1&1\\1&4\end{vmatrix}=4-1=3,$$ $$\begin{vmatrix}x&1\\1&4\end{vmatrix}=4x-1,$$ $$\begin{vmatrix}x&1\\1&1\end{vmatrix}=x-1.$$ Therefore, $$\det = 3 - x(4x-1) + (x-1).$$ Simplify: $$\det = 3-4x^2+x+x-1 = 2-4x^2+2x.$$ So, $$2-4x^2+2x=2.$$ Hence, $$-4x^2+2x=0$$ $$2x(1-2x)=0.$$ Thus, $$x=0 \quad \text{or} \quad x=\frac12.$$ But $\vec u$ and $\vec v$ are **not perpendicular**, so $\vec u\cdot\vec v\ne 0$. Therefore, $$\vec u\cdot\vec v=\frac12.$$ --- 4. **Find $|3\vec u+5\vec v|$** Use $$|3\vec u+5\vec v|^2=(3\vec u+5\vec v)\cdot(3\vec u+5\vec v).$$ So, $$|3\vec u+5\vec v|^2=9|\vec u|^2+25|\vec v|^2+30(\vec u\cdot\vec v).$$ Substitute $$|\vec u|=|\vec v|=1, \qquad \vec u\cdot\vec v=\frac12:$$ $$|3\vec u+5\vec v|^2=9+25+30\cdot \frac12=34+15=49.$$ Thus, $$|3\vec u+5\vec v|=7.$$ --- 5. **Final answer** $$\boxed{7}$$ The derived answer matches the stored correct answer.

More from Vector Algebra

View written solutionFree

Correct answer: 7

Given data

Let $|\,\vec u\,|=1,\qquad |\,\vec v\,|=1$ and $\vec u,\vec v$ are not perpendicular.

Also, $\vec u\cdot \vec w=1,\qquad \vec v\cdot \vec w=1,\qquad \vec w\cdot \vec w=4$ so $|\vec w|^2=4 \implies |\vec w|=2.$

The volume of the parallelepiped formed by $\vec u,\vec v,\vec w$ is $|\vec u\cdot(\vec v\times \vec w)|=\sqrt2.$

We need to find $|3\vec u+5\vec v|.$

Use Gram determinant for volume

For vectors $\vec u,\vec v,\vec w$ , the square of the volume is the determinant of the Gram matrix:

\vec u\cdot\vec u & \vec u\cdot\vec v & \vec u\cdot\vec w\\ \vec v\cdot\vec u & \vec v\cdot\vec v & \vec v\cdot\vec w\\ \vec w\cdot\vec u & \vec w\cdot\vec v & \vec w\cdot\vec w \end{pmatrix}.$$ Let $$\vec u\cdot\vec v=x.$$ Then the Gram matrix becomes $$\begin{pmatrix} 1 & x & 1\\ x & 1 & 1\\ 1 & 1 & 4 \end{pmatrix}.$$ Since $V=\sqrt2$, we have $$V^2=2.$$ So, $$\det\begin{pmatrix} 1 & x & 1\\ x & 1 & 1\\ 1 & 1 & 4 \end{pmatrix}=2.$$ --- 3. **Compute the determinant** Expand along the first row: $$\det = 1\begin{vmatrix}1&1\\1&4\end{vmatrix}-x\begin{vmatrix}x&1\\1&4\end{vmatrix}+1\begin{vmatrix}x&1\\1&1\end{vmatrix}.$$ Now, $$\begin{vmatrix}1&1\\1&4\end{vmatrix}=4-1=3,$$ $$\begin{vmatrix}x&1\\1&4\end{vmatrix}=4x-1,$$ $$\begin{vmatrix}x&1\\1&1\end{vmatrix}=x-1.$$ Therefore, $$\det = 3 - x(4x-1) + (x-1).$$ Simplify: $$\det = 3-4x^2+x+x-1 = 2-4x^2+2x.$$ So, $$2-4x^2+2x=2.$$ Hence, $$-4x^2+2x=0$$ $$2x(1-2x)=0.$$ Thus, $$x=0 \quad \text{or} \quad x=\frac12.$$ But $\vec u$ and $\vec v$ are **not perpendicular**, so $\vec u\cdot\vec v\ne 0$. Therefore, $$\vec u\cdot\vec v=\frac12.$$ --- 4. **Find $|3\vec u+5\vec v|$** Use $$|3\vec u+5\vec v|^2=(3\vec u+5\vec v)\cdot(3\vec u+5\vec v).$$ So, $$|3\vec u+5\vec v|^2=9|\vec u|^2+25|\vec v|^2+30(\vec u\cdot\vec v).$$ Substitute $$|\vec u|=|\vec v|=1, \qquad \vec u\cdot\vec v=\frac12:$$ $$|3\vec u+5\vec v|^2=9+25+30\cdot \frac12=34+15=49.$$ Thus, $$|3\vec u+5\vec v|=7.$$ --- 5. **Final answer** $$\boxed{7}$$ The derived answer matches the stored correct answer.

Vector Algebra question

Vector Algebra question

More from Vector Algebra

Vector Algebra question

Vector Algebra question

More from Vector Algebra