Interactive OP Malhotra ISC Class 12 Solutions Chapter 21 Vectors Chapter Test engage students in active learning and exploration.
S Chand Class 12 ICSE Maths Solutions Chapter 21 Vectors Chapter Test
Question 1.
Find the magnitude of the vector \(\vec{a}\) = 2 \(\hat{i}\) – 6 \(\hat{j}\) – 3 k.
Answer:
Given
\(\vec{a}\) = 2 \(\hat{i}\) – 6 \(\hat{j}\) – 3 \(\hat{k}\)
∴ |\(\vec{a}\)| = \(\sqrt{2^2+(-6)^2+(-3)^2}\)
= \(\sqrt{4+36+9}\)
= 7
Question 2.
Find the direction ratios and the direction cosines of the vector \(\vec{r}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\).
Answer:
∴ Direction numbers of ratios of vector \(\vec{r}\) be < 2, 3, 1 > Thus D’ cosines of \(\vec{r}\) are
< \(\frac{2}{\sqrt{2^2+3^2+1^2}}\), \(\frac{3}{\sqrt{2^2+3^2+1^2}}\), \(\frac{1}{\sqrt{2^2+3^2+1^2}}\) >
i.e. < \(\frac{2}{\sqrt{14}}\), \(\frac{3}{\sqrt{14}}\), \(\frac{1}{\sqrt{14}}\) >
Question 3.
(i) Find the position vector of a point R which divides the line joining the points P(\(\hat{i}\) + 2 \(\hat{j}\) – \(\hat{k}\)) and Q(\(-\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) in the ratio 2 : 1
(i) internally and
(ii) externally.
(iii) Find the position vector of a point which divides the join of points with position vector \(\vec{a}\) – 2 \(\vec{b}\) and 2 \(\vec{a}\) + \(\vec{b}\) externally in the ratio 2 : 1.
Answer:
Question 4.
Find the value of p for which p(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) is a unit vector.
Answer:
Let \(\vec{a}\) = p(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\))
∴|\(\vec{a}\)| = \(\sqrt{p^2+p^2+p^2}\) = \(\sqrt{3}\) p
Since \(\vec{a}\) is a unit vector
∴ |\(\vec{a}\)| = 1
⇒ |\(\vec{a}\)|^2 = 1
⇒ (\(\sqrt{3}\) p)^2 = 1
⇒ 3 p2 = 1
⇒ p = ± \(\frac{1}{\sqrt{3}}\)
Question 5.
Write the value of p for which \(\vec{a}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 9 \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + p \(\hat{j}\) + 3 \(\hat{k}\) are parallel.
Answer:
Given \(\vec{a}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 9 \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + p \(\hat{j}\) + 3 \(\hat{k}\)
Since \( \vec{a}\) || \(\vec{b}\)
∴ \(\vec{a}\) = λ \(\vec{b}\) for some non-zero scalar λ
⇒ (3 \(\hat{i}\) + 2 \(\hat{j}\) + 9 \(\hat{k}\)) = λ(\(\hat{i}\) + p \(\hat{j}\) + 3 \(\hat{k}\))
Comparing the coefficients of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) on both sides, we have 3 = λ
and 2 = λp
⇒ 2 = 3 p
⇒ p = \(\frac{2}{3}\)
Question 6.
Find the position vector of the mid-point of the line segment A B, where A is the point (3, 4, -2) and B is the point (1, 2, 4).
Answer:
Given P.V of A = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 2 \(\hat{k}\)
and P.V of B = \(\hat{i}\) + 2 \(\hat{j}\) + 4 \(\hat{k}\)
∴ P.V of mid point of line segment AB
= \(\frac{P . V \text { of } A + P . V \text { of } B}{2}\)
= \(\frac{(3 \hat{i}+4 \hat{j}-2 \hat{k})+(\hat{i}+2 \hat{j}+4 \hat{k})}{2} \)
= \(\frac{4 \hat{i}+6 \hat{j}+2 \hat{k}}{2}\)
= 2 \(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\)
Question 7.
Write a vector of magnitude (i) 15 units in the direction of vector \(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\).
(ii) 7 units in the direction \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\).
Answer:
(i) Let \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\)
∴ |\(\vec{a}\)| = \(\sqrt{1^2+(-2)^2+2^2}\) = 3
∴ unit vector in the direction of \(\vec{a}\)
= \(\hat{a}\) = \(\frac{\vec{a}}{|\vec{a}|}\)
= \(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\)
Thus required vector of magnitude 15 in the direction of \(\vec{a}\)
= 15 \(\hat{a}\)
= 5(\(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\))
= 5 \(\hat{i}\) – 10 \(\hat{j}\) + 10 \(\hat{k}\)
(ii) Given \(\vec{a}\) = \(\hat{i}\) – 2 \(\hat{j}\)
∴ unit vector in the direction of \(\vec{a}\)
= \(\hat{a}\) = \(\frac{\vec{a}}{|\vec{a}|}\)
= \(\frac{\hat{i}-2 \hat{j}}{\sqrt{1^2+(-2)^2}}\)
= \(\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}\)
= \(\frac{1}{\sqrt{5}}\) \(\hat{i}\) – \(\frac{2}{\sqrt{5}}\) \(\hat{j}\)
Hence the required vector of magnitude 7 units in the direction of \(\vec{a}\)
= 7 \(\hat{a}\) = 7(\(\frac{1}{\sqrt{5}}\) \(\hat{i}\) – \(\frac{2}{\sqrt{5}}\) \(\hat{j}\))
= \(\frac{7 \sqrt{5}}{5}\) \(\hat{i}\) – \(\frac{14 \sqrt{5}}{5}\) \(\hat{j}\)
Question 8.
If \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\vec{b}\) = 4 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\) and \(\vec{c}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\), find a vector of magnitude 6 units which is parallel to the vector \(\overrightarrow{2 a}\) – \(\vec{b}\) + \(\overrightarrow{3 c}\).
Answer:
Given \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\);
\(\vec{b}\) = 4 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\) and
\(\vec{c}\) = \(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\)
∴ 2 \(\vec{a}\) – \(\vec{b}\) + 3 \(\vec{c}\)
= 2(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)) – (4 \(\hat{i}\) – 2 \(\hat{j}\) + 3 \(\hat{k}\)) + 3(\(\hat{i}\) – 2 \(\hat{j}\) + \(\hat{k}\))
= \(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\)
|2 \(\vec{a}\) – \(\vec{b}\) + 3 \(\vec{c}\)|
= \(\sqrt{1^2+(-2)^2+2^2}\) = 3
Thus unit vector in the direction of 2 \(\vec{a}\) – \(\vec{b}\) + 3 \(\vec{c}\)
= \(\frac{2 \vec{a}-\vec{b}+3 \vec{c}}{|2 \vec{a}-\vec{b}+3 \vec{c}|}\)
= \(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\)
Hence the required vector of magnitude 6 units in the direction of 2 \(\vec{a}\) – \(\vec{b}\) + 3 \(\vec{c}\)
= 6 \(\frac{(2 \vec{a}-\vec{b}+3 \vec{c})}{|2 \vec{a}-\vec{b}+3 \vec{c}|}\)
= 6 (\(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\))
= 2 \(\hat{i}\) – 4 \(\hat{j}\) + 4 \(\hat{k}\)
Question 9.
Show that the points A(-2 i + 3 j + 5 k), B(i + 2 \(\hat{j}\) + 3 \(\hat{k}\)) and C(7 \(\hat{i}\) – \(\hat{k}\)) are collinear.
Answer:
Given
P.V of A = -2 \(\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\)
P.V of B = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)
and
P.V of C = 7 \(\hat{i}\) – \(\hat{k}\)
Here \(\overrightarrow{A B}\) = P . V of B – P . V of A
= \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\) – (-2 \(\hat{i}\) + 3 \(\hat{j}\) + 5 \(\hat{k}\))
= 3 \(\hat{i}\) – \(\hat{j}\) – 2 \(\hat{k}\)
\(\overrightarrow{\mathrm{BC}}\) = P.V. of C – P.V of B
= 7 \(\hat{i}\) – \(\hat{k}\) – (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\))
= 6 \(\hat{i}\) – 2 \(\hat{j}\) – 4 \(\hat{k}\)
⇒ \(\overrightarrow{\mathrm{BC}}\) = 2(3 \(\hat{i}\) – \(\hat{j}\) – 2 \(\hat{k}\))
= 2 \(\overrightarrow{\mathrm{AB}}\)
Thus \(\overrightarrow{\mathrm{BC}}\) is parallel to \(\overrightarrow{\mathrm{AB}}\) and B is common to both vectors \(\overrightarrow{\mathrm{BC}}\) and \(\overrightarrow{\mathrm{AB}}\). Thus A, B and C are collinear points.
Question 10.
Show that the points A, B and C with position vectors, \(\vec{a}\) = 3 \(\hat{i}\) – 4 \(\hat{j}\) – 4 \(\hat{k}\), \(\vec{b}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) and \(\vec{c}\) = \(\hat{i}\) – 3 \(\hat{j}\) – 5 \(\hat{k}\) respectively form the vertices of a right angled triangle.
Answer:
