Class 12 RD Sharma Solutions – Chapter 25 Vector or Cross Product – Exercise 25.1 | Set 3

Question 25. If $|\vec{a}|=\sqrt{26}$ , $|\vec{b}|= 7$ and $|\vec{a}\times\vec{b}|=35$ , find $\vec{a}.\vec{b}$

Solution:

We know that,
=> $\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n}$
=> $|\vec{a}\times\vec{b} |= |\vec{a}||\vec{b}|\sin\theta\|hat{n}|$
=> $35 = \sqrt{26}\times7|\sin\theta|\times1$
=> $\sin\theta = \dfrac{35}{7\sqrt{26}}$
=> $\sin\theta = \dfrac{5}{\sqrt{26}}$
As $\cos^2\theta + \sin^2\theta =1$ ,
=> $\cos\theta = \sqrt{1-\sin^2\theta}$
=> $\cos\theta = \sqrt{1-(\dfrac{5}{\sqrt{26}})^2}$
=> $\cos\theta = \sqrt{1-\dfrac{25}{26}}$
=> $\cos\theta = \dfrac{1}{\sqrt{26}}$
Thus,
=> $\vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta$
=> $\vec{a}.\vec{b} = 7\sqrt{26}\times\dfrac{1}{\sqrt{26}}$
=> $\vec{a}.\vec{b} = 7$

Question 26. Find the area of the triangle formed by O, A, B when $\vec{OA} = \hat{i}+2\hat{j}+3\hat{k}$ , $\vec{OB} = -3\hat{i}-2\hat{j}+\hat{k}$

Solution:

The area of a triangle whose adjacent sides are given by $\vec{a}$ and $\vec{b}$ is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$
=> $\vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}$
=> $\vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\-3 & -2 & 1\end{vmatrix}$
=> $\vec{OA}\times\vec{OB} = \hat{i}[2+6]-\hat{j}[1+9]+\hat{k}[-2+6]$
=> $\vec{OA}\times\vec{OB} = 8\hat{i}-10\hat{j}+4\hat{k}$
=> Area = $\dfrac{1}{2} |\vec{OA}\times\vec{OB}|$
=> Area = $\dfrac{1}{2}\sqrt{8^2+(-10^2)+4^2}$
=> Area = $\dfrac{1}{2}\sqrt{64+100+16}$
=> Area = $\dfrac{1}{2}\times6\sqrt{45}$
=> Area = $3\sqrt{5}$ square units.

Question 27. Let $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ , $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\vec{c} = 2\hat{i}-\hat{j}+4\hat{k}$ . Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c}.\vec{d} = 15$

Solution:

Given that $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ .
=> $\vec{d}.\vec{a} =0$ ……….(1)
=> $\vec{d}.\vec{b} =0$ ……….(2)
Also,
=> $\vec{c}.\vec{d} = 15$ …….(3)
Let $\vec{d} = d_1\hat{i}+d_2\hat{j}+d_3\hat{k}$
From eq(1),
=> d₁ + 4d₂ + 2d₃ = 0
From eq(2),
=> 3d₁ – 2d₂ + 7d₃ = 0
From eq(3),
=> 2d₁ – d₂ + 4d₃ = 15
On solving the 3 equations we get,
d₁= 160/3, d₂= -5/3, and d₃= -70/3,
=> $\vec{d} = \dfrac {1}{3}(160\hat{i}-5\hat{j}-70\hat{k})$

Question 28. Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , where $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}-2\hat{k}$ .

Solution:

Given that, $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}-2\hat{k}$
Let $\vec{c} = \vec{a}+\vec{b}$
=> $\vec{c} = (3\hat{i}+2\hat{j}+2\hat{k})+ (\hat{i}+2\hat{j}-2\hat{k})$
=> $\vec{c} = \hat{i}[3+1] +\hat{j}[2+2] +\hat{k}[2-2]$
=> $\vec{c} = 4\hat{i} +4\hat{j}$
Let $\vec{d} = \vec{a}-\vec{b}$
=> $\vec{d} = (3\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-2\hat{k})$
=> $\vec{d} = \hat{i}[3-1] +\hat{j}[2-2] +\hat{k}[2+2]$
=> $\vec{d} = 2\hat{i} +4\hat{k}$
A vector perpendicular to both $\vec{c}$ and $\vec{d}$ is,
=> $\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\c_1 & c_2 & c_3\\d_1 & d_2 & d_3\end{vmatrix}$
=> $\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4 & 4 & 0\\2 & 0 & 4\end{vmatrix}$
=> $\vec{c}\times\vec{d} = \hat{i}[16-0]-\hat{j}[16-0]+\hat{k}[0-8]$
=> $\vec{c}\times\vec{d} = 16\hat{i}-16\hat{j}-8\hat{k}$
To find the unit vector,
=> $\hat{p} = \dfrac{\vec{c}\times\vec{d}}{|\vec{c}\times\vec{d}|}$
=> $\hat{p} = \dfrac{1}{\sqrt{16^2+(-16)^2+(-8)^2}}(16\hat{i}-16\hat{j}-8\hat{k})$
=> $\hat{p} = \dfrac{1}{\sqrt{256+256+64}}(16\hat{i}-16\hat{j}-8\hat{k})$
=> $\hat{p} = \dfrac{1}{24}(16\hat{i}-16\hat{j}-8\hat{k})$
=> $\hat{p} = \dfrac{1}{3}(2\hat{i}-2\hat{j}-\hat{k})$

Question 29. Using vectors, find the area of the triangle with the vertices A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8).

Solution:

Given, A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8)
Let,
=> $\vec{a} = A = 2\hat{i}+3\hat{j}+5\hat{k}$
=> $\vec{b} = B = 3\hat{i}+5\hat{j}+8\hat{k}$
=> $\vec{c} = C = 2\hat{2}+7\hat{j}+8\hat{k}$
Then,
=> $\vec{AB} = \vec{b}-\vec{a}$
=> $\vec{AB} = (3\hat{i}+5\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})$
=> $\vec{AB} = \hat{i}[3-2]+\hat{j}[5-3]+\hat{k}[8-5]$
=> $\vec{AB} = \hat{i}+2\hat{j}+3\hat{k}$
=> $\vec{AC} = \vec{c}-\vec{a}$
=> $\vec{AC} = (2\hat{2}+7\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})$
=> $\vec{AC} = \hat{i}[2-2]+\hat{j}[7-3]+\hat{k}[8-5]$
=> $\vec{AC} = 4\hat{j}+3\hat{k}$
The area of a triangle whose adjacent sides are given by $\vec{a}$ and $\vec{b}$ is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$
=> $\vec{AB}\times\vec{AC}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2& 3\\0 & 4 & 3\end{vmatrix}$
=> $\vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]$
=> $\vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}$
=> Area = $\dfrac{1}{2}|\vec{AB}\times\vec{AC}|$
=> Area = $\dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}$
=> Area = √61/2

Question 30. If $\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ , $\vec{b}=-\hat{i}+\hat{k}$ , $\vec{c}=2\hat{j}-\hat{k}$ are three vectors, find the area of the parallelogram having diagonals $(\vec{a}+\vec{b})$ and $(\vec{b}+\vec{c})$ .

Solution:

Given, $\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ , $\vec{b}=-\hat{i}+\hat{k}$ , $\vec{c}=2\hat{j}-\hat{k}$
Let,
=> $\vec{c} = (\vec{a}+\vec{b})$
=> $\vec{c} = (2\hat{i}-3\hat{j}+\hat{k})+(-\hat{i}+\hat{k})$
=> $\vec{c} = (2-1)\hat{i}+(-3)\hat{j}+(1+1)\hat{k}$
=> $\vec{c} = \hat{i}-3\hat{j}+2\hat{k}$
=> $\vec{d} = (\vec{b}+\vec{c})$
=> $\vec{d} = (-\hat{i}+\hat{k})+(2\hat{j}-\hat{k})$
=> $\vec{d} = -\hat{i}+2\hat{j}$
The area of the parallelogram having diagonals $\vec{c}$ and $\vec{d}$ is $\dfrac{1}{2}|\vec{c}\times\vec{d}|$
=> $\vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3& 2\\-1 & 2 & 0\end{vmatrix}$
=> $\vec{c}\times\vec{d} = \hat{i}[0-4] -\hat{j}[0+2]+\hat{k}[2-3]$
=> $\vec{c}\times\vec{d} = -4\hat{i}-2\hat{j}-\hat{k}$
=> Area = $\dfrac{1}{2}|\vec{c}\times\vec{d}|$
=> Area = $\dfrac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2}$
=> Area = $\dfrac{1}{2}\sqrt{21}$
=> Area = √21/2

Question 31. The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}+5\hat{k}$ and $\hat{i}-2\hat{j}-3\hat{k}$ . Find the unit vector parallel to one of its diagonals. Also, find its area.

Solution:

Given a parallelogram ABCD and its 2 sides AB and BC.
By triangle law of addition,
=> $\vec{AC} = \vec{AB}+\vec{BC}$
=> $\vec{AC} = (2\hat{i}-4\hat{j}+5\hat{k})+(\hat{i}-2\hat{j}-3\hat{k})$
=> $\vec{AC} = \hat{i}[2+1] +\hat{j}[-4-2]+\hat{k}[5-3]$
=> $\vec{AC} = 3\hat{i}-6\hat{j}+2\hat{k}$
Unit vector is,
=> $\hat{p} = \dfrac{\vec{AC}}{|\vec{AC}|}$
=> $\hat{p} = \dfrac{1}{\sqrt{3^2+(-6)^2+2^2}}(3\hat{i}-6\hat{j}+2\hat{k})$
=> $\hat{p} = \dfrac{1}{\sqrt{49}}(3\hat{i}-6\hat{j}+2\hat{k})$
=> $\hat{p} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})$
Area of a parallelogram whose adjacent sides are given is $|\vec{a}\times\vec{b}|$
=> $|\vec{AB}\times\vec{BC}| = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2 & -4 & 5\\1 & -2 & -3\end{vmatrix}$
=> $|\vec{AB}\times\vec{BC}| = \hat{i}[12+10]-\hat{j}[-6-5]+\hat{k}[-4+4]$
=> $|\vec{AB}\times\vec{BC}| = 22\hat{i}+11\hat{j}$
Thus area is,
=> Area = $|22\hat{i}+11\hat{j}|$
=> Area = $\sqrt{22^2+11^2}$
=> Area = $\sqrt{605}$
=> Area = 11 √5 square units

Question 32. If either $\vec{a}=0$ or $\vec{b}=0$ , then $\vec{a}\times\vec{b}=\vec{0}$ . Is the converse true? Justify with example.

Solution:

Let us take two parallel non-zero vectors $\vec{a}$ and $\vec{b}$
=> $\vec{a}\times\vec{b} = \vec{0}$
For example,
$\vec{a} = \hat{i}$ and $\vec{b}=2\hat{i}$
=> $\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 0 & 0\\2 & 0 & 0\end{vmatrix}$
=> $\vec{a}\times\vec{b} = 0$
But,
=> $|\vec{a}| = \sqrt{1^2} =1$
=> $|\vec{b}| = \sqrt{2^2} =2$
Hence the converse may not be true.

Question 33. If $\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ , $\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ and $\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ , then verify that $\vec{a}\times(\vec{b}\times\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$ .

Solution:

Given, $\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ , $\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ and $\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$
=> $(\vec{b}+\vec{c}) = (b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}$
=> $\vec{a}\times(\vec{b}\times\vec{c}) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\(b_1+c_1) & (b_2+c_2) & (b_3+c_3)\end{vmatrix}$
=> $\vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2(b_3+c_3)-a_3(b_2+c_2)]-\hat{j}[a_1(b_3+c_3)-a_3(b_1+c_1)]+\hat{k}[a_1(b_2+c_2)-a_2(b_1+c_1)]$
=> $\vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]$ …..eq(1)
Now,
=> $\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}$
=> $\vec{a}\times\vec{b} = \hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]$
And,
=> $\vec{a}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\c_1 & c_2 & c_3\end{vmatrix}$
=> $\vec{a}\times\vec{c} = \hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2]$
Thus,
=> $\vec{a}\times\vec{b}+\vec{a}\times\vec{c} = (\hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]) + (\hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2])$
=> $\vec{a}\times\vec{b}+\vec{a}\times\vec{c} = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]$ …eq(2)
Thus eq(1) = eq(2)
Hence proved.

Question 34(i). Using vectors find the area of the triangle with the vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).

Solution:

Given, A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5)
=> $\vec{a} = A = \hat{i}+\hat{j}+2\hat{k}$
=> $\vec{b} = B = 2\hat{i}+3\hat{j}+5\hat{k}$
=> $\vec{c} = C = \hat{i}+5\hat{j}+5\hat{k}$
Now 2 sides of the triangle are given by,
=> $\vec{AB} = \vec{b}-\vec{a}$
=> $\vec{AB} = (2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})$
=> $\vec{AB} = \hat{i}[2-1] +\hat{j}[3-1]+\hat{j}[5-2]$
=> $\vec{AB} = \hat{i}+2\hat{j}+3\hat{k}$
=> $\vec{AC} = \vec{c}-\vec{a}$
=> $\vec{AC} = (\hat{i}+5\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})$
=> $\vec{AC} = \hat{i}[1-1] +\hat{j}[5-1]+\hat{j}[5-2]$
=> $\vec{AC} = 4\hat{j}+3\hat{k}$
Area of the triangle whose adjacent sides are given is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$
=> $\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\0 & 4 & 3\end{vmatrix}$
=> $\vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]$
=> $\vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}$
Thus area of the triangle is,
=> Area = $\dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}$
=> Area = $\dfrac{1}{2}\sqrt{36+9+16}$
=> Area = √61/2

Question 34(ii). Using vectors find the area of the triangle with the vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1).

Solution:

Given, A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1)
=> $\vec{a} = A = \hat{i}+2\hat{j}+3\hat{k}$
=> $\vec{b} = B = 2\hat{i}-1\hat{j}+4\hat{k}$
=> $\vec{c} = C = 4\hat{i}+5\hat{j}-1\hat{k}$
Now 2 sides of the triangle are given by,
=> $\vec{AB} = \vec{b}-\vec{a}$
=> $\vec{AB} = (2\hat{i}-1\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}-1\hat{k})$
=> $\vec{AB} = \hat{i}[2-1] +\hat{j}[-1-2]+\hat{j}[4-3]$
=> $\vec{AB} = \hat{i}-3\hat{j}+\hat{k}$
=> $\vec{AC} = \vec{c}-\vec{a}$
=> $\vec{AC} = (4\hat{i}+5\hat{j}-1\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})$
=> $\vec{AC} = \hat{i}[4-1] +\hat{j}[5-2]+\hat{j}[-1-3]$
=> $\vec{AC} = 3\hat{i}+3\hat{j}-4\hat{k}$
Area of the triangle whose adjacent sides are given is $\dfrac{1}{2}|\vec{a}\times\vec{b}|$
=> $\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3 & 1\\3 & 3 & -4\end{vmatrix}$
=> $\vec{AB}\times\vec{AC} = \hat{i}[12-3]-\hat{j}[-4-3]+\hat{k}[3+9]$
=> $\vec{AB}\times\vec{AC} = 9\hat{i}+7\hat{j}+12\hat{k}$
Thus area of the triangle is,
=> Area = $\dfrac{1}{2}\sqrt{(9)^2+(7)^2+12^2}$
=> Area = $\dfrac{1}{2}\sqrt{81+49+144}$
=> Area = √274/2

Question 35. Find all the vectors of magnitude $10\sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+4\hat{k}$ .

Solution:

Given, $\vec{a} = \hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}+4\hat{k}$
A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is,
=> $\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 1\\-1 & 3 & 4\end{vmatrix}$
=> $\vec{a}\times\vec{b} = \hat{i}[8-3]-\hat{j}[4+1]+\hat{k}[3+2]$
=> $\vec{a}\times\vec{b} = 5\hat{i}-5\hat{j}+5\hat{k}$
Unit vector is,
=> $\hat{p} = \dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$
=> $\hat{p} = \dfrac{1}{\sqrt{5^2+(-5)^2+5^2}}(5\hat{i}-5\hat{j}+5\hat{k})$
=> $\hat{p} = \dfrac{1}{5\sqrt{3}}(5\hat{i}-5\hat{j}+5\hat{k})$
=> $\hat{p} = \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$
Now vectors of magnitude $10\sqrt{3}$ are given by,
=> $10\sqrt{3}\hat{p} = \pm10\sqrt{3}\times \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$
=> Required vectors, $\pm10(\hat{i}-\hat{j}+\hat{k})$

Question 36. The adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}-5\hat{k}$ and $2\hat{i}+2\hat{j}+3\hat{k}$ . Find the 2 unit vectors parallel to its diagonals. Also, find its area of the parallelogram.

Solution:

Given, $\vec{AB}=2\hat{i}-4\hat{j}-5\hat{k}$ and $\vec{BC} = 2\hat{i}+2\hat{j}+3\hat{k}$
=> $\vec{AC} = \vec{AB} + \vec{BC}$
=> $\vec{AC} = (2\hat{i}-4\hat{j}-5\hat{k})+(2\hat{i}+2\hat{j}+3\hat{k})$
=> $\vec{AC} = 4\hat{i}-\hat{j}-2\hat{k}$
Unit vector is,
=> $\hat{p} = \dfrac{\vec{AC}}{|\vec{AC}|}$
=> $\hat{p} = \dfrac{1}{\sqrt{4^2+(-1)^2+(-2)^2}}(4\hat{i}-\hat{j}-2\hat{k})$
=> $\hat{p} = \dfrac{1}{\sqrt{21}}(4\hat{i}-\hat{j}-2\hat{k})$
Area is given by $|\vec{AB}\times\vec{BC}|$ ,

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Last Updated : 28 Mar, 2021

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Class 12 RD Sharma Solutions – Chapter 25 Vector or Cross Product – Exercise 25.1 | Set 3

Question 25. If $|\vec{a}|=\sqrt{26}$ , $|\vec{b}|= 7$ and $|\vec{a}\times\vec{b}|=35$ , find $\vec{a}.\vec{b}$

Question 26. Find the area of the triangle formed by O, A, B when $\vec{OA} = \hat{i}+2\hat{j}+3\hat{k}$ , $\vec{OB} = -3\hat{i}-2\hat{j}+\hat{k}$

Question 27. Let $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ , $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\vec{c} = 2\hat{i}-\hat{j}+4\hat{k}$ . Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c}.\vec{d} = 15$

Question 28. Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , where $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}-2\hat{k}$ .

Question 29. Using vectors, find the area of the triangle with the vertices A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8).

Question 30. If $\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ , $\vec{b}=-\hat{i}+\hat{k}$ , $\vec{c}=2\hat{j}-\hat{k}$ are three vectors, find the area of the parallelogram having diagonals $(\vec{a}+\vec{b})$ and $(\vec{b}+\vec{c})$ .

Question 31. The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}+5\hat{k}$ and $\hat{i}-2\hat{j}-3\hat{k}$ . Find the unit vector parallel to one of its diagonals. Also, find its area.

Question 32. If either $\vec{a}=0$ or $\vec{b}=0$ , then $\vec{a}\times\vec{b}=\vec{0}$ . Is the converse true? Justify with example.

Question 33. If $\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ , $\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ and $\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ , then verify that $\vec{a}\times(\vec{b}\times\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$ .

Question 34(i). Using vectors find the area of the triangle with the vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).

Question 34(ii). Using vectors find the area of the triangle with the vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1).

Question 35. Find all the vectors of magnitude $10\sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+4\hat{k}$ .

Question 36. The adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}-5\hat{k}$ and $2\hat{i}+2\hat{j}+3\hat{k}$ . Find the 2 unit vectors parallel to its diagonals. Also, find its area of the parallelogram.

CBSE Class 12 Computer Science

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Class 12 RD Sharma Solutions – Chapter 25 Vector or Cross Product – Exercise 25.1 | Set 3

Question 25. If, and , find

Question 26. Find the area of the triangle formed by O, A, B when,

Question 27. Let , and. Find a vector which is perpendicular to bothand and

Question 28. Find a unit vector perpendicular to each of the vectors and , where and .

Question 29. Using vectors, find the area of the triangle with the vertices A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8).

Question 30. If , , are three vectors, find the area of the parallelogram having diagonals and .

Question 31. The two adjacent sides of a parallelogram are and . Find the unit vector parallel to one of its diagonals. Also, find its area.

Question 32. If either or , then . Is the converse true? Justify with example.

Question 33. If , and, then verify that .

Question 34(i). Using vectors find the area of the triangle with the vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).

Question 34(ii). Using vectors find the area of the triangle with the vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1).

Question 35. Find all the vectors of magnitude that are perpendicular to the plane of and .

Question 36. The adjacent sides of a parallelogram are and . Find the 2 unit vectors parallel to its diagonals. Also, find its area of the parallelogram.

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CBSE Class 12 Computer Science

GATE CS & IT 2024

Complete Machine Learning & Data Science Program

Question 25. If $|\vec{a}|=\sqrt{26}$ , $|\vec{b}|= 7$ and $|\vec{a}\times\vec{b}|=35$ , find $\vec{a}.\vec{b}$

Question 26. Find the area of the triangle formed by O, A, B when $\vec{OA} = \hat{i}+2\hat{j}+3\hat{k}$ , $\vec{OB} = -3\hat{i}-2\hat{j}+\hat{k}$

Question 27. Let $\vec{a}=\hat{i}+4\hat{j}+2\hat{k}$ , $\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}$ and $\vec{c} = 2\hat{i}-\hat{j}+4\hat{k}$ . Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c}.\vec{d} = 15$

Question 28. Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ , where $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b} = \hat{i}+2\hat{j}-2\hat{k}$ .

Question 30. If $\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ , $\vec{b}=-\hat{i}+\hat{k}$ , $\vec{c}=2\hat{j}-\hat{k}$ are three vectors, find the area of the parallelogram having diagonals $(\vec{a}+\vec{b})$ and $(\vec{b}+\vec{c})$ .

Question 31. The two adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}+5\hat{k}$ and $\hat{i}-2\hat{j}-3\hat{k}$ . Find the unit vector parallel to one of its diagonals. Also, find its area.

Question 32. If either $\vec{a}=0$ or $\vec{b}=0$ , then $\vec{a}\times\vec{b}=\vec{0}$ . Is the converse true? Justify with example.

Question 33. If $\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ , $\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ and $\vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ , then verify that $\vec{a}\times(\vec{b}\times\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$ .

Question 35. Find all the vectors of magnitude $10\sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+4\hat{k}$ .

Question 36. The adjacent sides of a parallelogram are $2\hat{i}-4\hat{j}-5\hat{k}$ and $2\hat{i}+2\hat{j}+3\hat{k}$ . Find the 2 unit vectors parallel to its diagonals. Also, find its area of the parallelogram.